# Quantitative Techniques For Management

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#### Quantitative Techniques For Management

THEORETICAL PROBABILITY DISTRIBUTIONS

Usual manager is forced to make decisions when there is uncertainty as to what will happen after the decisions are made. In this situation the mathematical theory of probability furnishes a tool that can be of great help to the decision maker. A probability function is a rule that assigns probabilities to each element of a set of events that may occur. Probability distribution can either discrete or continuous. A discrete probability distribution is sometimes called a probability mass function and a continuous one is called a probability density function.

PROBABILITY DISTRIBUTION

The probability distribution of a discrete random variable is the list of all possible values and the probability of each value occurring. This definition cannot be applied to continuous random variables because the probability of any one outcome is zero (there is an infinite number of outcomes and any finite number divided by infinity is zero). However, it is possible to list the probabilities of ranges of values. For discrete variables the probability distribution is called the probability mass function. For continuous variables, it is called the probability density function.

The knowledge of the theoretical probability distribution is of great use in the understanding and analysis of a large number of business and economic situations. For example, with the use of probability distribution, it is possible to test a hypothesis about a population, to take decision in the face of uncertainty, to make forecast, etc.

Theoretical probability distributions can be divided into two broad categories, viz. discrete and continuous probability distributions, depending upon whether the random variable is discrete or continuous. Although, there are a large number of distributions in each category, we shall discuss only some of them having important business and economic applications.

BINOMIAL DISTRIBUTION

Binomial distribution is a theoretical probability distribution which was given by James Bernoulli. The binomial distribution is used when there are exactly two mutually exclusive outcomes of a trial. These outcomes are appropriately labeled "success" and "failure". The binomial distribution is used to obtain the probability of observing x successes in N trials, with the probability of success on a single trial denoted by p. The binomial distribution assumes that p is fixed for all trials.

The sequence of trials under the above assumptions is also termed as Bernoulli Trials.

Probability Function or Probability Mass Function

Let n be the total number of repeated trials, p be the probability of a success in a trial and q be the probability of its failure so that q = 1 - p.

Let r be a random variable which denotes the number of successes in n trials. The possible values of r are 0, 1, 2, ...... n. We are interested in finding the probability of r successes out of n trials, i.e., P(r).

To find this probability, we assume that the first r trials are successes and remaining n -r trials are failures. Since different trials are assumed to be independent, the probability of this sequence is

Since out of n trials any r trials can be success, the number of sequences showing any r trials as success and remaining (n - r) trials as failure is nCr, where the probability of r successes in each trial is prqn-r. Hence, the required probability is P(r) nCrprqn r, where r = 0, 1, 2, ...... n.

Writing this distribution in a tabular form, we have

It should be noted here that the probabilities obtained for various values of r are the terms in the binomial expansion of (q + p)n and thus, the distribution is termed as Binomial Distribution. P (r)= nCrp r q n-r  is termed as the probability function or probability mass function (p.m.f.) of the distribution.

Summary Measures of Binomial Distribution

Fitting of Binomial Distribution

We often want to compare a set of data from observations with a theoretical probability distribution. Can the data be represented satisfactorily by a theoretical distribution? If so, the data can be represented very succinctly by the parameters of the theoretical distribution. Specifically, let us consider whether a set of data can be represented by a binomial distribution. The binomial distribution has two parameters, n and p.

In any practical case we will already know n, the number of trials. How can we estimate p, the probability of “success” in a single trial? An intuitive answer is that we can estimate p by the fraction of all the trials which were “successes,” that is, the proportion or relative frequency of “success.” It is possible to show mathematically that this intuitive answer is correct, an unbiased estimate of the parameter p.

Example: The following data give the number of seeds germinating (X) out of 10 on damp filter for 80 sets of seed. Fit a binomial distribution to the data.

Solution: Here the random variable X denotes the number of seeds germinating out of a set of 10 seeds. The total number of trials n = 10.

The mean of the given data

X= 0*6+1*20+2*28+3*12 4*8+5*6/80=174/80=2.175

Since mean of a binomial distribution is np, ∴ np = 2.175. Thus, we get.

p= 2.175/10 = 0.22 (approx.). Further, q = 1 - 0.22 = 0.78.

Using these values, we can compute P(X) = 10Cx (0.22) x (0.78) 10-x and then expected frequency [= N × P(X)] for X = 0, 1, 2, ...... 10. The calculated probabilities and the respective expected frequencies are shown in the following table:

Features of Binomial Distribution

1. It is a discrete probability distribution.
2. It depends upon two parameters n and p. It may be pointed out that a distribution is known if the values of its parameters are known.
3. The total number of possible values of the random variable are n + 1. The successive binomial coefficients are nC0, nC1, nC2 , .... nCn. Further, since nCr, nCn r , these coefficients are symmetric.

The values of these coefficients, for various values of n, can be obtained directly by using Pascal's triangle.

We can note that it is very easy to write this triangle. In the first row, both the coefficients will be unity because 1C0 = 1C1 . To write the second row, we write 1 in the beginning and the end and the value of the middle coefficients is obtained by adding the coefficients of the first row. Other rows of the Pascal's triangle can be written in a similar way.

1. The shape and location of binomial distribution changes as the value of p changes for a given value of n. It can be shown that for a given value of n, if p is increased gradually in the interval (0, 0.5), the distribution changes from a positively skewed to a symmetrical shape. When p = 0.5, the distribution is perfectly symmetrical. Further, for larger values of p the distribution tends to become more and more negatively skewed.
2. For a given value of p, which is neither too small nor too large, the distribution becomes more and more symmetrical as n becomes larger and larger.

Uses of Binomial Distribution

Binomial distribution is often used in various decision-making situations in business. Acceptance sampling plan, a technique of quality control, is based on this distribution. With the use of sampling plan, it is possible to accept or reject a lot of items either at the stage of its manufacture or at the stage of its purchase.

HYPERGEOMETRIC DISTRIBUTION

The hypergeometric distributionarises when a random selection (without repetition) is made among objects oftwo distinct types. Typical examples:

• Choose a team of 8 from a group of 10 boys and 7 girls
• Choose a committee of five from the legislature consisting of 52 Democrats and 48 Republicans

Letthere be a finite population of size N, where each item can be classified aseither a success or a failure. Let there be k successes in the population. If arandom sample of size n is taken from this population, then the probability ofr successes is given by

Example: A retailer has 10 identical televisionsets of a company out which 4 are defective. If 3 televisions are selected atrandom, construct the probability distribution of the number of defectivetelevision sets.

Solution: Let the random variable r denote thenumber of defective televisions. In terms of notations, we can write N = 10, k= 4 and n = 3.

Thus,we can write
P(r)= 4Cr* 6C3-r/ 10C3, r= 0,1,2,3

Thedistribution of r is hypergeometric. This distribution can also be written in atabular form as given below :

Binomial Approximation to Hypergeometric Distribution

The Hypergeometric distribution recognises the fact that we are samplingfrom a finite population without replacement, so that the result of a sample isdependent on the samples that have gone before it. Now imagine that thepopulation is very large, so that removing a sample of size n has nodiscernible effect on the population.

Then the probability that an individualsample will have the characteristic of interest is essentially constant and hasthe value D/M, because the probability of resampling an item in thepopulation, were one to replace items after sampling, would be very small. Insuch cases, the Hypergeometric distribution can be approximated by a Binomialas follows:

Hypergeometric(n, D, M) >> Binomial(n,D/M)

The rule most often quoted is that this approximation works well when n< 0.1 M.

Example : Thereare 200 identical radios out of which 80 are defective. If 5 radios areselected at random, construct the probability distribution of the number ofdefective radios by using (i) hypergeometric distribution and (ii) binomialdistribution.

Solution:

(i) It is given that N = 200, k = 80 and n = 5.
Let r be a hypergeometric random variable which denotes the number of defective radios,then

P(r)= 80Cr* 120C5-r/ 200C5, r=0,1,2,3,4,5

The probabilities for various values of r are given in the following table:

(ii) To use binomial distribution, we find p = 80/200 = 0.4.

P(r)   5Cr(0.4)r (0.6) 5 r .r    0,1, 2, 3,4,5

The probabilities forvarious values of r are given in the following table:

We note that these probabilities are in close conformity with the hypergeometric probabilities.

PASCAL DISTRIBUTION

In binomial distribution, we derived the probability mass function of the number of successes in n (fixed) Bernoulli trials. We can also derive the probability mass function of the number of Bernoulli trials needed to get r (fixed) successes. This distribution is known as Pascal distribution. Here r and p become parameters while n becomes a random variable.

We may note that r successes can be obtained in r or more trials i.e. possible values of the random  variable are r, (r + 1), (r + 2), ...... etc. Further, if n trials are required to get r successes, the nth trial must be a success. Thus, we can write the probability mass function of Pascal distribution as follows:

It can be shown that the mean and variance of Pascal distribution are r/p and rq/p2respectively. This distribution is also known as Negative Binomial Distribution because various values of P(n) are given by the terms of the binomial expansion of pr(1 - q)- r

GEOMETRICAL DISTRIBUTION

When r = 1, the Pascal distribution can be written as

P(n) = n-1 C0 pq n-1 = Pq n-1 where n = 1,2,3,.....

Here n is a random variable which denotes the number of trials required to get a success. This distribution is known as geometrical distribution. The mean and variance of the distribution are 1/p and q/p2 respectively.

UNIFORM DISTRIBUTION (DISCRETE RANDOM VARIABLE)

A discrete random variable is said to follow a uniform distribution if it takes various discrete values with equal probabilities.

If a random variable X takes values X1, X2, ...... Xn each with probability 1/n, the distribution of X is said to be uniform.

Exercise with Hints

1. The probability that a secretary will not put the correct postage on a letter is 0.20. What is the probability that this secretary will not put the correct postage:
(i) On 3 of 9 letters? (ii) On at least 3 of 9 letters? (iii) On at the most 3 of 9 letters?
Hint: Use binomial distribution.
2. (a) The mean of a binomial distribution is 4 and its standard deviation is √3 . What are the values of n, p and q?
(b) The mean and variance of a binomial distribution are 3 and 2 respectively. Find the probability that the variate takes values (i) less than or equal to 2 (ii) greater than or equal to 7.
Hint: Mean = np and Variance = npq.
3. (a) The probability of a man hitting a target is 1/4
(i) If he fires 7 times, what is the probability of his hitting the target at least twice? (ii) How many times
must he fire so that the probability of his hitting the target at least once is greater than 2/3?
(b) How many dice must be thrown so that there is better than even chance of obtaining at least one six?
Hint: (a) (ii) Probability of hitting the target at least once in n trials is 1- (¾)n
Find n such that this value is greater than 2/3. (b) Find n so that 1- (5/6)n> ½
4. A machine produces an average of 20% defective bolts. A batch is accepted if a sample of 5 bolts taken from the batch contains no defective and rejected if the sample contains 3 or more defectives. In other cases, a second sample is taken. What is the probability that the second sample is required?
Hint: A second sample is required if the first sample is neither rejected nor accepted.
5. A multiple choice test consists of 8 questions with 3 answers to each question (of which only one is correct). A student answers each question by throwing a balanced die and checking the first answer if he gets 1 or 2, the second answer if he gets 3 or 4 and the third answer if he gets 5 or 6. To get a distinction, the student must secure at least 75% correct answers. If there is no negative marking, what is the probability that the student secures a distinction?
Hint: He should attempt at least 6 questions.
6. What is the most probable number of times an ace will appear if a die is tossed (i) 50 times, (ii) 53 times?
Hint: Find mode.
7. Out of 320 families with 5 children each, what percentage would be expected to have (i) 2 boys and 3 girls, (ii) at least one boy, (iii) at the most one girl? Assume equal probability for boys and girls.

Hint: Multiply probability by 100 to obtain percentage.

8. Fit a binomial distribution to the following data:

Hint: See example
9. A question paper contains 6 questions of equal value divided into two sections of three questions each. If each question poses the same amount of difficulty to Mr. X, an examinee, and he has only 50% chance of solving it correctly, find the answer to the following :
(i) If Mr. X is required to answer only three questions from any one of the two sections, find the probability that he will solve all the three questions correctly.
(ii) If Mr. X is given the option to answer the three questions by selecting one question out of the two standing at serial number one in the two sections, one question out of the two standing at serial number two in the two sections and one question out of the two standing at serial number three in the two sections, find the probability that he will solve all three questions correctly.
Hint: (i) A section can be selected in 2C1 ways and the probability of attempting all thethree questions correctly is3c3(½)3. (ii) The probability of attempting correctly,one question out of two is2c1(½)2

A binomial random variable satisfies the relation 9P(X = 4) = P(X = 2) for n = 6. Find the value of the parameter p.

Hint: P (X = 2) = 6C2p2q4

10. Three fair coins are tossed 3,000 times. Find the frequency distributions of the number of heads and tails and tabulate the results. Also calculate mean and standard deviation of each distribution.
Hint: See example .
11. Take 100 sets of 10 tosses of an unbiased coin. In how many cases do you expect to get (i) 6 heads and 4 tails and (ii) at least 9 heads?
Hint: Use binomial distribution with n = 10 and p = 0.5.
12. In a binomial distribution consisting of 5 independent trials, the probabilities of 1 and 2 successes are 0.4096 and 0.2048 respectively. Find the probability of success.
Hint: Use the condition P(1) = 2P(2).
13. For a binomial distribution, the mean and variance are respectively 4 and 3. Calculate the probability of getting a nonzero value of this distribution.
Hint: Find P(r ¹ 0).
14. (a) There are 300, seemingly identical, tyres with a dealer. The probability of a tyre being defective is 0.3. If 2 tyres are selected at random, find the probability that there is non defective tyre.
(b) If instead of 300 tyres the dealer had only 10 tyres out of which 3 are defective, find the probability that no tyre is defective in a random sample of 2 tyres.
(c) Write down the probability distribution of the number of defectives in each case.
Hint: Use (a) binomial (b) hypergeometric distributions.
15. Write down the mean and variance of a binomial distribution with parameters n and p. If the mean and variance are 4 and 8/3 respectively, find the values of n and p. State whether it is symmetric for these values?
Hint: Binomial distribution is symmetric when p = 0.5.
16. Evaluate k if f(x) = k, when x = 1, 2, 3, 4, 5, 6 and = 0 elsewhere, is a probability mass function. Find its mean and standard deviation.
Hint: Σ f (x) =1.
17. If a die is thrown 6 times, calculate the probability that :
(i) a score of 3 or less occurs on exactly 2 throws;
(ii) a score of more than 2 occurs on exactly 3 throws;
(iii) a score of 5 or less occurs at least once;
(iv) a score of 2 or less occurs on at least 5 throws.
Hint: Use binomial distribution with n = 6 and a different value of p in each case.
18. If we take 1,280 sets each of 10 tosses of a fair coins, in how many sets should we expect to get 7 heads and 3 tails?
Hint: See example .
19. If a production unit is made up from 20 identical components and each component has a probability of 0.25 of being defective, what is the average number of defective components in a unit? Further, what is the probability that in a unit (i) less than 3 components are defective? (ii) exactly 3 components are defective?
Hint: Take n = 20 and p = 0.25.
20. It is known from the past experience that 80% of the students in a school do their home work. Find the probability that during a random check of 10 students
(i) all have done their home work,
(ii) at the most 2 have not done their home work,
(iii) at least one has not done the home work.
Hint: Take n = 10 and p = 0.8.
21. There are 24 battery cells in a box containing 6 defective cells that are randomly mixed. A customer buys 3 cells. What is the probability that he gets one defective cell?
Hint: Use hypergeometric distribution.
22. There are 400 tyres in the stock of a wholesaler among which 40 tyres, having slight defects, are randomly mixed. A retailer purchases 6 tyres from this stock. What is the probability that he gets at least 4 non defective tyres?
Hint: n is less than 5% of N.

POISSON DISTRIBUTION

This distribution was derived by a noted mathematician, Simon D. Poisson, in 1837. In probability theory, the Poisson distribution (or Poisson law of small numbers) is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time and/or space if these events occur with a known average rate and independently of the time since the last event. (The Poisson distribution can also be used for the number of events in other specified intervals such as distance, area or volume.)

Poisson Process

Let us assume that on an average 3 telephone calls are received per 10 minutes at a telephone exchange desk and we want to find the probability of receiving a telephone call in the next 10 minutes. In an effort to apply binomial distribution, we can divide the interval of 10 minutes into 10 intervals of 1 minute each so that the probability of receiving a telephone call (i.e., a success) in each minute (i.e., trial) becomes 3/10 ( note that p = m/n, where m denotes mean).

Thus, there are 10 trials which are independent, each with probability of success = 3/10. However, the main difficulty with this formulation is that, strictly speaking, these trials are not Bernoulli trials. One essential requirement of such trials, that each trial must result into one of the two possible outcomes, is violated here. In the above example, a trial, i.e. an interval of one minute, may result into 0, 1, 2...... successes depending upon whether the exchange desk receives none, one, two, ...... telephone calls respectively.

One possible way out is to divide the time interval of 10 minutes into a large number of small intervals so that the probability of receiving two or more telephone calls in an interval becomes almost zero. This is illustrated by the following table which shows that the probabilities of receiving two calls decreases sharply as the number of intervals are increased, keeping the average number of calls, 3 calls in 10 minutes in our example, as constant.

Using symbols, we may note that as n increases then p automatically declines in such a way that the mean m (= np) is always equal to a constant. Such a process is termed as a Poisson Process. The chief characteristics of Poisson process can be summarized as given below:

1. The number of occurrences in an interval is independent of the number of occurrences in another interval.
2. The expected number of occurrences in an interval is constant.
3. It is possible to identify a small interval so that the occurrence of more than one event, in any interval of this size, becomes extremely unlikely.

Probability Mass Function

The probability mass function (p.m.f.) of Poisson distribution can be derived as a limit of p.m.f. of binomial distribution when n→∞such that m (= np) remains constant. Thus, we can write

&there4 Β= µ3223= m2/ m3=1/m

Since m is a positive quantity, therefore, Β1 is always positive and hence the Poisson distribution is always positively skewed. We note that Β1→0 as µ ® ¥, therefore the distribution tends to become more and more symmetrical for large values of m.

Further,Β2= µ4/ µ22=m+3m2/m2= 3 + 1/m → 3 as m → ∞.This result shows that the distribution becomes normal for large values of m.

(d) Mode: As in binomial distribution, a Poisson variate r will be mode

if P(r -1) ≤ P(r )≥ P(r +1) The inequality P(r -1) ≤ P(r ) can be written as

Similarly, the inequality P(r ) ≥ P(r +1) can be shown to simply that r ≥ m - 1 .... (2) Combining (1) and (2), we can write m - 1 ≤ r ≤m.

Case I: When m is not an integer

The integral part of m will be mode.

Case II: When m is an integer

The distribution is bimodal with values m and m - 1.

Example: The average number of customer arrivals per minute at a super bazaar is 2. Find the probability that during one particular minute (i) exactly 3 customers will arrive, (ii) at the most two customers will arrive, (iii) at least one customer will arrive.

Solution: It is given that m = 2. Let the number of arrivals per minute be denoted by the random variable r. The required probability is given by

Example : An executive makes, on an average, 5 telephone calls per hour at a cost which may be taken as Rs 2 per call. Determine the probability that in any hour the telephone calls' cost (i) exceeds Rs 6, (ii) remains less than Rs 10.

Solution: The number of telephone calls per hour is a random variable with mean = 5.

The required probability is given by

Example : A company makes electric toys. The probability that an electric toy is defective is 0.01. What is the probability that a shipment of 300 toys will contain exactly 5 defectives?
Solution: Since n is large and p is small, Poisson distribution is applicable. The random variable is the number of defective toys with mean m = np = 300 × 0.01 = 3.

The required probability is given by

P(r=5)= e-3.35/5!= 0.04979*243/120=0.10082

Example : In a town, on an average 10 accidents occur in a span of 50 days. Assuming that the number of accidents per day follow Poisson distribution, find the probability that there will be three or more accidents in a day.

Solution: The random variable denotes the number accidents per day. Thus, we have. m=10/50 = 0.2. The required probability is given by

P(r≥3)=1-P(r≤2)=1-e-0.2[1+0.2+(0.2)2/2!]=1-0.8187*1.22=0.00119

Example : A car hire firm has two cars which it hires out every day. The number of demands for a car on each day is distributed as a Poisson variate with mean 1.5. Calculate the proportion of days on which neither car is used and the proportion of days on which some demand is refused. [ e-1.5 = 0.2231]

Solution: When both car are not used, r = 0

P(r = 0) = e-1.5 = 0.2231. Hence the proportion of days on which neither car is used is 22.31%.
Further, some demand is refused when more than 2 cars are demanded, i.e., r > 2

Hence the proportion of days is 19.13%.

Example : A firm produces articles of which 0.1 percent are usually defective. It packs them in cases each containing 500 articles. If a wholesaler purchases 100 such cases, how many cases are expected to be free of defective items and how many are expected to contain one defective item?

Solution: The Poisson variate is number of defective items with mean

m= 1/1000 x 500 = 0.5

Probability that a case is free of defective items

P(r = 0) = e-0.5 = 0.6065.Hence the number of cases having no defective items = 0.6065 × 100 = 60.65

Similarly, P(r =1) = e-0.5 *0.5 = 0.6065*0.5 = 0.3033.Hence the number of cases having one defective item are 30.33.

Example : A manager accepts the work submitted by his typist only when there is no mistake in the work. The typist has to type on an average 20 letters per day of about 200 words each. Find the chance of her making a mistake (i) if less than 1% of the letters submitted by her are rejected; (ii) if on 90% of days all the work submitted by her is accepted. [As the probability of making a mistake is small, you may use Poisson distribution. Take e = 2.72].

Solution: Let p be the probability of making a mistake in typing a word.

(i) Let the random variable r denote the number of mistakes per letter. Since 20 letters are typed, r will follow Poisson distribution with mean = 20p. Since less than 1% of the letters are rejected, it implies that the probability of making at least one mistake is less than 0.01, i.e., P(r ≥ 1) ≤ 0.01 or 1 - P(r = 0) ≤ 0.01

P(r ≥ 1) ≤ 0.01

or

1 - P(r = 0) ≤0.01

⇒=1 - e-20p≤0.01

or e-20p≤0.99

Taking log of both sides

`– 20p.log 2.72 ≤ log 0.99 - (20* 0.4346)p ≥ 1.9956`

No. of mistakes per page  : 0      1    2     3

Frequency : 211   90  19    5

– 8.692p ≤ - 0.0044 or p≥0.0044/8.692=0.00051.

(ii) In this case r is a Poisson variate which denotes the number of mistakes per day. Since the typist has to type 20 × 200 = 4000 words per day, the mean number of mistakes = 4000p.
It is given that there is no mistake on 90% of the days, i.e.,

P(r = 0) = 0.90 or e-4000p = 0.90

Taking log of both sides, we have

- 4000p log 2.72 = log 0.90 or - 4000*0.4346p = 1.9542 = - 0.0458

p= 0.0458/4000*0.4346 = 0.000026

Example : A manufacturer of pins knows that on an average 5% of his product is defective. He sells pins in boxes of 100 and guarantees that not more than 4 pins will be defective. What is the probability that the box will meet the guaranteed quality?

Solution: The number of defective pins in a box is a Poisson variate with mean equal to 5. A box will meet the guaranteed quality if r ≤ 4. Thus, the required probability is given by

Poisson Approximation to Binomial

The probability mass function for the Poisson distribution can be derived from the Binomial probability mass function by making n extremely large while p becomes very small, but within the constraint that np remains finite. When n, the number of trials becomes large, the computation of probabilities by using the binomial probability mass function becomes a cumbersome task. Usually, when n ≥ 20 and p ≤ 0.05, Poisson distribution can be used as an approximation to binomial with parameter m = np.

Example : Find the probability of 4 successes in 30 trials by using (i) binomial distribution and (ii) Poisson distribution. The probability of success in each trial is given to be 0.02.

solution:

(i) Here n = 30 and p = 0.02
∴ P(r=4)= 30C4(0.02)4(0.98)26

= 27405*0.00000016*0.59= 0.00259.

(ii) Here m = np = 30 × 0.02 = 0.6
∴ P(r=4)= e-.6(0.6)4/4!= 0.548*0.1296/24=0.00296

Fitting of a Poisson Distribution

To fit a Poisson distribution to a given frequency distribution, we first compute its mean m. Then the probabilities of various values of the random variable r are computed by using the probability mass function P(r) = e-m mr. These probabilities are then multiplied by N, the total frequency, to get expected frequencies.

Example : The following mistakes per page were observed in a book:

No. of mistakes per page: 0      1   2  3

Frequency                    : 211 90 19  5

Fit a Poisson distribution to find the theoretical frequencies.

Solution: The mean of the given frequency distribution is

m= 0*211+1*90+2*19+3*5/211+90+19+5=143/325=0.44

Calculation of theoretical (or expected) frequencies

We can write P(r) = e-0.44 (0.44)r/ r!. Substituting r = 0, 1, 2 and 3, we get the probabilities for various values of r, as shown in the following table:

Features of Poisson Distribution

1. It is discrete probability distribution.
2. It has only one parameter m.
3. The range of the random variable is 0 ≤ r < ∞.
4. The Poisson distribution is a positively skewed distribution. The skewness decreases as m increases.

Uses of Poisson Distribution

1. This distribution is applicable to situations where the number of trials is large and the probability of a success in a trial is very small.
2. It serves as a reasonably good approximation to binomial distribution when n ≥ 20 and p ≤ 0.05.

Exercise with Hints

1. If 2% of the electric bulbs manufactured by a company are defective, find the probability that in a sample of 200 bulbs (i) less than 2 bulbs are defective, (ii) more than 3 bulbs are defective. (Given e- 4 = 0.0183).
Hint:  m= 2/100 x 200 = 4
2. If r is a Poisson variate such that P(r) = P(r + 1), what are the mean and standard deviation of r?
Hint: Find m by using the given condition.
3. The number of arrivals of telephone calls at a switch board follows a Poisson process at an average rate of 8 calls per 10 minutes. The operator leaves for a 5 minutes tea break. Find the probability that (a) at the most two calls go unanswered and (b) 3 calls go unanswered, while the operator is away.
Hint: m = 4.
4. What probability model is appropriate to describe a situation where 100 misprints are distributed randomly throughout the 100 pages of a book? For this model, what is the probability that a page observed at random will contain (i) no misprint, (ii) at the most two misprints, (iii) at least three misprints?
Hint: The average number of misprint per page is unity.
5. If the probability of getting a defective transistor in a consignment is 0.01, find the mean and standard deviation of the number of defective transistors in a large consignment of 900 transistors. What is the probability that there is at the most one defective transistor in the consignment?
Hint: The average number of transistors in a consignment is 900 × 0.01.
6. In a certain factory turning out blades, there is a small chance 1/500 for any one blade to be defective. The blades are supplied in packets of 10. Use Poisson distribution to compute the approximate number of packets containing no defective, one defective, two defective, three defective blades respectively in a consignment of 10,000 packets.
Hint: The random variable is the number of defective blades in a packet of 10 blades.
7. A manufacturer knows that 0.3% of items produced in his factory are defective. If the items are supplied in boxes, each containing 250 items, what is the probability that a box contains (i) no defective, (ii) at the most two defective items?
Hint: m= 0.3/100 X 250 = 0.75
8. A random variable r follows Poisson distribution, where P(r = 2) = P(r = 3). Find (i) P(r = 0), (ii) P(1 ≤ r ≤ 3).
Hint: P(1 ≤ r ≤ 3) = P(r = 1) + P(r = 2) + P(r = 3).
9. If X is a Poisson variate such that P(X = 2) = 9P(X = 4) + 90P(X = 6), find the mean and variance of X.
Hint: mean = Variance.
10. Lots of 400 wall-clocks are purchased by a retailer. The retailer inspects sample of 20 clocks from each lot and returns the lot to the supplier if there are more than two defectives in the sample. Suppose a lot containing 30 defective clocks is received by the retailer, what is the probability that it will be returned to the supplier?
Hint: n = 20 and p = 30/400.
11. An industrial area has power breakdown once in 15 days, on the average. Assuming that the number of breakdowns follow a Poisson process, what is the probability of (i) no power breakdown in the next six days, (ii) more than one power breakdown in the next six days?
Hint: The random variable is the number of power breakdowns in six days.
12. After correcting the proofs of first 50 pages or so of a book, it is found that on the average there are 3 errors per 5 pages. Use Poisson probabilities and estimate the number of pages with 0, 1, 2, 3, errors in the whole book of 1,000 pages. [Given that e- 0.6 = 0.5488].
Hint: Take random variable as the number of errors per page.
13. Between 2 and 4 p.m., the number of phone calls coming into the switch board of a company is 300. Find the probability that during one particular minute there will be (i) no phone call at all, (ii) exactly 3 calls, (iii) at least 7 calls. [Given e- 2 = 0.13534 and e-0.5 = 0.60650].
Hint: Random variable is the number of calls per minute.
14. It is known that 0.5% of ball pen refills produced by a factory are defective. These refills are dispatched in packagings of equal numbers. Using Poisson distribution determine the number of refills in a packing to be sure that at least 95% of them contain no defective refills.
Hint: Let n be the number of refills in a package, then m = 0.005n.
15. Records show that the probability is 0.00002 that a car will have a flat tyre while driving over a certain bridge. Find the probability that out of 20,000 cars driven over the bridge, not more than one will have a flat tyre.
Hint: The random variable is number of cars driven over the bridge having flat tyre.
16. A radioactive source emits on the average 2.5 particles per second. Calculate the probability that two or more particles will be emitted in an interval of 4 seconds.
Hint: m = 2.5 * 4.
17. The number of accidents in a year attributed to taxi drivers in a city follows Poisson distribution with mean 3. Out of 1,000 taxi drivers, find approximately the number of drivers with (i) no accident in a year, (ii) more than 3 accidents in a year.
[Given e-1 = 0.3679, e- 2 = 0.1353, e- 3 = 0.0498].
Hint: Number of drivers = probability × 1000.
18. A big industrial plant has to be shut down for repairs on an average of 3 times in a month. When more than 5 shut downs occur for repairs in a month, the production schedule cannot be attained. Find the probability that production schedule cannot be attained in a given month, assuming that the number of shut downs are a Poisson variate.
Hint: Find P(r 5).
19. A manager receives an average of 12 telephone calls per 8-hour day. Assuming that the number of telephone calls received by him follow a Poisson variate, what is the probability that he will not be interrupted by a call during a meeting lasting 2 hours?
Hint: Take m = 3.
20. Assuming that the probability of a fatal accident in a factory during a year is 1/1200, calculate the probability that in a factory employing 300 workers, there will be at least two fatal accidents in a year. [Given e- 0.25 = 0.7788].
Hint: The average number of accidents per year in the factory = 0.25.
21. If 2% of electric bulbs manufactured by a certain company are defective, find the probability that in a sample of 200 bulbs (i) less than 2 bulbs are defective (ii) more than 3 bulbs are defective.
[Given e-4 = 0.0183].
Hint: m = 4.
22. If for a Poisson variate X, P(X = 1) = P(X = 2), find P(X = 1 or 2). Also find its mean and standard deviation.
Hint: Find m from the given condition.
23. If 5% of the families in Calcutta do not use gas as a fuel, what will be the probability of selecting 10 families in a random sample of 100 families who do not use gas as a fuel? You may assume Poisson distribution. [Given e-5 = 0.0067].
Hint: m = 5, find P(r = 10).
24. The probability that a Poisson variate X takes a positive value is 1 - e-1.5. Find the variance and also the probability that X lies between –1.5 and 1.5.
Hint: 1- e-1.5 = P(r > 0). Find P(-1.5 < X < 1.5) = P(X = 0) + P(X = 1).
25. 250 passengers have made reservations for a flight from Delhi to Mumbai. If the probability that a passenger, who has reservation, will not turn up is 0.016, find the probability that at the most 3 passengers will not turn up.
Hint: The number of passengers who do not turn up is a Poisson variate.

EXPONENTIAL DISTRIBUTION

The random variable in case of Poisson distribution is of the type; the number of arrivals of customers per unit of time or the number of defects per unit length of cloth, etc. Alternatively, it is possible to define a random variable, in the context of Poisson Process, as the length of time between the arrivals of two consecutive customers or the length of cloth between two consecutive defects, etc. The probability distribution of such a random variable is termed as Exponential Distribution.

Since the length of time or distance is a continuous random variable, therefore exponential distribution is a continuous probability distribution.

Probability Density Function

The probability density function of a continuous random variable is a function which can be integrated to obtain the probability that the random variable takes a value in a given interval. Let t be a random variable which denotes the length of time or distance between the occurrence of two consecutive events or the occurrence of the first event and m be the average number of times the event occurs per unit of time or length.

Further, let A be the event that the time of occurrence between two consecutive events or the occurrence of the first event is less than or equal to t and f(t) and F(t) denote the probability density function and the distribution (or cumulative density) function of t respectively.

We can write P(A) + P(Ā) =1 or F (t) + P(Ā) = 1. Note that, by definition,F (t) = P(A) Further, P(Ā) is the probability that the length of time between the occurrence of two consecutive events or the occurrence of first event is greater than t. This is also equal to the probability that no event occurs in the time interval t. Since the mean number of occurrence of events in time t is mt, we have , by Poisson distribution

Example : A telephone operator attends on an average 150 telephone calls per hour. Assuming that the distribution of time between consecutive calls follows an exponential distribution, find the probability that (i) the time between two consecutive calls is less than 2 minutes, (ii) the next call will be received only after 3 minutes.

Solution: Here m = the average number of calls per minute = 150/60= 2.5.

Example : The average number of accidents in an industry during a year is estimated to be 5. If the distribution of time between two consecutive accidents is known to be exponential, find the probability that there will be no accidents during the next two months

solution

Here m denotes the average number of accidents per month = 5/12
&there4 P(t>2)= 1- F(2) = e-(5/12)*2 = e-0.833 = 0.4347

Example: The distribution of life, in hours, of a bulb is known to be exponential with mean life of 600 hours. What is the probability that (i) it will not last more than 500 hours, (ii) it will last more than 700 hours?

solution

Since the random variable denote hours, therefore m=1/600

(i) P(t ≤500) = F(500)
= 1- e(-1/600)*500 = 1-e -0.833
= 0.5653

(ii) P(t > 700) = 1 - F(700)= e(700/600)= e(-1.1667)= 0.3114

UNIFORM DISTRIBUTION (CONTINUOUS VARIABLE)

A continuous random variable X is said to be uniformly distributed in a close interval (a, b) with probability density function p(X)if P(X ) = 1/Β-Α for Α≤X≤Β and = 0 Otherwise The Uniform distribution is alternatively known as rectangular distribution.The diagram of the probability density function is shown in the figure.

Example : The buses on a certain route run after every 20 minutes. If a person arrives at the bus stop at random, what is the probability that

(a) he has to wait between 5 to 15 minutes,

(b) he gets a bus within 10 minutes,

(c) he has to wait at least 15 minutes.

Solution: Let the random variable X denote the waiting time, which follows a uniform

NORMAL DISTRIBUTION

The normal distribution is a theoretical function commonly used in inferential statistics as an approximation to sampling distributions. In general, the normal distribution provides a good model for a random variable, when:

1. There is a strong tendency for the variable to take a central value;
2. Positive and negative deviations from this central value are equally likely;
3. The frequency of deviations falls off rapidly as the deviations become larger.

As an underlying mechanism that produces the normal distribution, we can think of an infinite number of independent random (binomial) events that bring about the values of a particular variable. For example, there are probably a nearly infinite number of factors that determine a person's height (thousands of genes, nutrition, diseases, etc.). Thus, height can be expected to be normally distributed in the population.

Since Gauss used this curve to describe the theory of accidental errors of measurements involved in the calculation of orbits of heavenly bodies, it is also called as Gaussian curve.

The Conditions of Normality

In order that the distribution of a random variable X is normal, the factors affecting its observations must satisfy the following conditions:

1. A large number of chance factors: The factors, affecting the observations of a random variable, should be numerous and equally probable so that the occurrence or non-occurrence of any one of them is not predictable.
2. Condition of homogeneity: The factors must be similar over the relevant population although, their incidence may vary from observation to observation.
3. Condition of independence: The factors, affecting observations, must act independently of each other.
4. Condition of symmetry: Various factors operate in such a way that the deviations of observations above and below mean are balanced with regard to their magnitude as well as their number.

Random variables observed in many phenomena related to economics, business and other social as well as physical sciences are often found to be distributed normally. For example, observations relating to the life of an electrical component, weight of packages, height of persons, income of the inhabitants of certain area, diameter of wire, etc., are affected by a large number of factors and hence, tend to follow a pattern that is very similar to the normal curve.

In addition to this, when the number of observations become large, a number of probability distributions like Binomial, Poisson, etc., can also be approximated by this distribution.

Probability Density Function

If X is a continuous random variable, distributed normally with mean m and standard deviation s , then its p.d.f. is given by

Here p and s are absolute constants with values 3.14159.... and 2.71828.... respectively. It may be noted here that this distribution is completely known if the values of mean m and standard deviation s are known. Thus, the distribution has two parameters, viz. mean and standard deviation.

Shape of Normal Probability Curve

For given values of the parameters, m and s, the shape of the curve corresponding to normal probability density function p(X) is as shown in Figure.

It should be noted here that although we seldom encounter variables that have a range from - ∞ to ∞, as shown by the normal curve, nevertheless the curves generated by the relative frequency histograms of various variables closely resembles the shape of normal curve.

Properties of Normal Probability Curve

A normal probability curve or normal curve has the following properties:

1. It is a bell shaped symmetrical curve about the ordinate at X = m . The ordinate is maximum at X = m .
2. It is unimodal curve and its tails extend infinitely in both directions, i.e., the curve is asymptotic to X axis in both directions.
3. All the three measures of central tendency coincide, i.e., mean = median = mode
4. The total area under the curve gives the total probability of the random variable taking values between -¥ to ¥ . Mathematically, it can be shown that

5. Since median = m, the ordinate at X = m divides the area under the normal curve into two equal parts, i.e.,

6. The value of p(X) is always non-negative for all values of X, i.e., the whole curve lies above X axis.
7. The points of inflexion (the point at which curvature changes) of the curve are at X = m ± s .
8. The quartiles are equidistant from median, i.e., Md - Q1 = Q3 - Md , by virtue of symmetry. Also Q1 = m - 0.6745 s , Q3 = m + 0.6745 s , quartile deviation = 0.6745 s and mean deviation = 0.8s , approximately.
9. Since the distribution is symmetrical, all odd ordered central moments are zero.
10. The successive even ordered central moments are related according to the following recurrence formula µ 2n = (2n - 1) σ 2µ 2n - 2 for = 1, 2, 3, ......
11. The value of moment coefficient of skewness Β1 is zero.
12. The coefficient of kurtosis

Β2= µ422= 3 σ44= 3

Note that the above expression makes use of property .

1. Area property: The area under the normal curve is distributed by its standard deviation in the following manner.

1. The area between the ordinates at m - sand m + sis 0.6826. This implies that for a normal distribution about 68% of the observations will lie betweenµ - σ and µ + s.
2. The area between the ordinates at µ– 2σand µ+ 2σis 0.9544. This implies that for a normal distribution about 95% of the observations will lie between µ– 2σand µ+ 2σ.
3. The area between the ordinates at µ– 3σand µ/em>+ 3σis 0.9974. This implies that for a normal distribution about 99% of the observations will lie between µ– 3σand µ+ 3σ. This result shows that, practically, the range of the distribution is 6σ although, theoretically, the range is from – ∞ to ∞.

Probability of Normal Variate in an Interval

Let X be a normal variate distributed with mean m and standard deviation s, also written in abbreviated form as X ~ N(m, s) The probability of X lying in the interval (X1, X2) is given by

In terms of figure, this probability is equal to the area under the normal curve between the ordinates at X = X1 and X = X2 respectively.

Note: It may be recalled that the probability that a continuous random variable takes a particular value is defined to be zero even though the event is not impossible.

It is obvious from the above that, to find P(X1≤X ≤ X2), we have to evaluate an integral which might be cumbersome and time consuming task. Fortunately, an alternative procedure is available for performing this task. To devise this procedure, we define a new variable z= X-µ/ σ.

Further, from the reproductive property, it follows that the distribution of z is also normal.

Thus, we conclude that if X is a normal variate with mean m and standard deviation s, then z= X-µ/σ  is a normal variate with mean zero and standard deviation unity. Since the parameters of the distribution of z are fixed, it is a known distribution and is termed as standard normal distribution (s.n.d.). Further, z is termed as a standard normal variate (s.n.v.).

It is obvious from the above that the distribution of any normal variate X can always be transformed into the distribution of standard normal variate z. This fact can be utilised to evaluate the integral given above.

In terms of figure, this probability is equal to the area under the standard normal curve between the ordinates at z = z1 and z = z2.

Since the distribution of z is fixed, the probabilities of z lying in various intervals are tabulated. These tables can be used to write down the desired probability.

Example: Using the table of areas under the standard normal curve, find the following probabilities :

(i) P(0 ≤ z ≤ 1.3)          (ii) P(–1 ≤ z ≤ 0)          (iii) P(–1 ≤ z ≤ 12)
(iv) P( z ≥ 1.54)            (v) P(|z| > 2)                (vi) P(|z| < 2)

Solution: The required probability, in each question, is indicated by the shaded are of the corresponding figure.

1. From the table, we can write P(0 ≤ z ≤ 1.3) = 0.4032.
2. We can write P(–1 ≤ z ≤ 0) = P(0 ≤ z ≤ 1), because the distribution is symmetrical.
3. We can write
P(z ≥ 1.54) = 0.5000 – P(0 ≤ z ≤ 1.54) = 0.5000 – 0.4382 = 0.0618.
4. P(|z| > 2) = P(z > 2) + P(z < – 2) = 2P(z > 2) = 2[0.5000 - P(0 ≤ z ≤ 2)]
= 1 – 2P(0 ≤ z ≤ 2) = 1 – 2 × 0.4772 = 0.0456.
(vi) P(|z| < 2) = P(- 2 ≤ z ≤ 0) + P(0 ≤ z ≤ 2) = 2P(0 ≤ z ≤ 2) = 2 × 0.4772 = 0.9544.

Example : Determine the value or values of z in each of the following situations:
(a) Area between 0 and z is 0.4495.
(b) Area between – ∞ to z is 0.1401.
(c) Area between – ∞ to z is 0.6103.
(d) Area between – 1.65 and z is 0.0173.
(e) Area between – 0.5 and z is 0.5376.

Solution:

(a) On locating the value of z corresponding to an entry of area 0.4495 in the table of areas under the normal curve, we have z = 1.64. We note that the same situation may correspond to a negative value of z. Thus, z can be 1.64 or - 1.64.

(b) Since the area between –∞ to z < 0.5, z will be negative. Further, the area between z and 0 = 0.5000 – 0.1401 = 0.3599. On locating the value of z corresponding to this entry in the table, we get z = –1.08.

(c) Since the area between –∞ to z > 0.5000, z will be positive. Further, the area between 0 to z = 0.6103 - 0.5000 = 0.1103. On locating the value of z corresponding to this entry in the table, we get z = 0.28.

(d) Since the area between –1.65 and z < the area between –1.65 and 0 (which, from table, is 0.4505), z is negative. Further z can be to the right or to the left of the value –1.65. Thus, when z lies to the right of –1.65, its value, corresponds to an area (0.4505 – 0.0173) = 0.4332, is given by z = –1.5 (from table). Further, when z lies to the left of - 1.65, its value, corresponds to an area (0.4505 + 0.0173) = 0.4678, is given by z = –1.85 (from table).

(e) Since the area between –0.5 to z > area between –0.5 to 0 (which, from table, is 0.1915), z is positive. The value of z, located corresponding to an area (0.5376 – 0.1915) = 0.3461, is given by 1.02.

Example : If X is a random variate which is distributed normally with mean 60 and standard deviation 5, find the probabilities of the following events:
(i) 60 ≤ X ≤ 70, (ii) 50 ≤ X ≤ 65, (iii) X > 45, (iv) X ≤ 50.

Solution: It is given that m = 60 and s = 5

(i) Given X1 = 60 and X2 = 70, we can write

z1 = X1-µ/σ = 60-60/5=0 and z2= X2-µ/σ= 70-60-5 = 2.

∴ P(60 ≤ X ≤ 70) = P(0 ≤ z ≤ 2) = 0.4772 (from table).

(ii) Here X1 = 50 and X2 = 65, therefore, we can write

z1= 50-60/5 = -2 and z2=65-60/5 = 1

Hence P(50 ≤ X ≤ 65) = P(–2 ≤ z ≤ 1) = P(0 ≤ z ≤ 2) + P(0 ≤ z ≤ 1)
= 0.4772 + 0.3413 = 0.8185

P(X >45) =  P(z ≥ =
45-60
5
) = P(z ≥-3)

= P(-3≤ z≤0)+ P(0≤z≤∞)= P(0≤z≤3)+P(0≤z≤∞)
=0.4987+0.5000=0.9987

Example : The average monthly sales of 5,000 firms are normally distributed with mean Rs 36,000 and standard deviation Rs 10,000. Find :
(i) The number of firms with sales of over Rs 40,000.
(ii) The percentage of firms with sales between Rs 38,500 and Rs 41,000.
(iii) The number of firms with sales between Rs 30,000 and Rs 40,000.

Solution: Let X be the normal variate which represents the monthly sales of a firm. Thus X ~ N(36,000, 10,000).

Example : In a large institution, 2.28% of employees have income below Rs 4,500 and 15.87% of employees have income above Rs. 7,500 per month. Assuming the distribution of income to be normal, find its mean and standard deviation.

Solution: Let the mean and standard deviation of the given distribution be and srespectively.

Example : Marks in an examination are approximately normally distributed with mean 75 and standard deviation 5. If the top 5% of the students get grade A and the bottom 25% get grade F, what mark is the lowest A and what mark is the highest F?

Solution: Let A be the lowest mark in grade A and F be the highest mark in grade F. From the given information, we can write

Example : The mean inside diameter of a sample of 200 washers produced by a machine is 5.02 mm and the standard deviation is 0.05 mm. The purpose for which these washers are intended allows a maximum tolerance in the diameter of 4.96 to 5.08 mm, otherwise the washers are considered as defective. Determine the percentage of defective washers produced by the machine on the assumption that diameters are normally distributed.

Example : The average number of units produced by a manufacturing concern per day is 355 with a standard deviation of 50. It makes a profit of Rs 1.50 per unit. Determine the percentage of days when its total profit per day is (i) between Rs 457.50 and Rs 645.00, (ii) greater than Rs 682.50 (assume the distribution to be normal). The area between z = 0 to z = 1 is 0.34134, the area between z = 0 to z = 1.5 is 0.43319 and the area between z = 0 to z = 2 is 0.47725, where z is a standard normal variate.

Solution: Let X denote the profit per day. The mean of X is 355*1.50 = Rs 532.50 and its S.D. is 50*1.50 = Rs 75. Thus, X ~ N (532.50, 75).

Example : The distribution of 1,000 examinees according to marks percentage is given below:

Assuming the marks percentage to follow a normal distribution, calculate the mean and standard deviation of marks. If not more than 300 examinees are to fail, what should be the passing marks?

Solution: Let X denote the percentage of marks and its mean and S.D. be mand s respectively. From the given table, we can write

Example : In a certain book, the frequency distribution of the number of words per page may be taken as approximately normal with mean 800 and standard deviation 50. If three pages are chosen at random, what is the probability that none of them has between 830 and 845 words each?

Solution: Let X be a normal variate which denotes the number of words per page. It is given that X ~ N(800, 50).

The probability that a page, select at random, does not have number of words between 830 and 845, is given by

Thus, the probability that none of the three pages, selected at random, have number of words lying between 830 and 845 = (0.91)3 = 0.7536.

Example : At a petrol station, the mean quantity of petrol sold to a vehicle is 20 litres per day with a standard deviation of 10 litres. If on a particular day, 100 vehicles took 25 or more litres of petrol, estimate the total number of vehicles who took petrol from the station on that day. Assume that the quantity of petrol taken from the station by a vehicle is a normal variate.

Solution: Let X denote the quantity of petrol taken by a vehicle. It is given that X ~ N(20, 10).

Normal Approximation to Binomial Distribution

Normal distribution can be used as an approximation to binomial distribution when n is large and neither p nor q is very small. If X denotes the number of successes with probability p of a success in each of the n trials, then X will be distributed approximately normally with mean np and standard deviation √ npq.

ICorrection for Continuity
Since the number of successes is a discrete variable, to use normal approximation, we have make corrections for continuity.

Example : An unbiased die is tossed 600 times. Use normal approximation to binomial to find the probability obtaining

(i) more than 125 aces,
(ii) number of aces between 80 and 110,
(iii) exactly 150 aces.

Solution: Let X denote the number of successes, i.e., the number of aces.

Normal/ Approximation to Poisson Distribution

Normal distribution can also be used to approximate a Poisson distribution when its parameter m ≥10. If X is a Poisson variate with mean m, then, for m ≥ 10, the distribution of X can be taken as approximately normal with mean m and standard deviation √so that z=x-m/√m  is a standard normal variate.

Example : A random variable X follows Poisson distribution with parameter 25. Use normal approximation to Poisson distribution to find the probability that X is greater than or equal to 30.

Solution:

Fitting a Normal Curve

A normal curve is fitted to the observed data with the following objectives:

1. To provide a visual device to judge whether it is a good fit or not.
2. Use to estimate the characteristics of the population.

The fitting of a normal curve can be done by

(a) The Method of Ordinates or
(b) The Method of Areas.

(a) Method of Ordinates: In this method, the ordinate f(X) of the normal curve, for various values of the random variate X are obtained by using the table of ordinates for a standard normal variate.

Example : Fit a normal curve to the following data :

Note: If the class intervals are not continuous, they should first be made so.

∴ µ= 45-10*(10/100)=44

(b) Method of Areas: Under this method, the probabilities or the areas of the random variable lying in various intervals are determined. These probabilities are then multiplied by N to get the expected frequencies. This procedure is explained below for the data of the above example.

Exercise with Hints

1. In a metropolitan city, there are on the average 10 fatal road accidents in a month (30 days). What is the probability that (i) there will be no fatal accident tomorrow,
(ii) next fatal accident will occur within a week?
Hint: Take m = 1/3 and apply exponential distribution.
2. A counter at a super bazaar can entertain on the average 20 customers per hour. What is the probability that the time taken to serve a particular customer will be
(i) less than 5 minutes, (ii) greater than 8 minutes?
Hint: Use exponential distribution.
3. The marks obtained in a certain examination follow normal distribution with mean 45 and standard deviation 10. If 1,000 students appeared at the examination, calculate the number of students scoring (i) less than 40 marks, (ii) more than 60 marks and (iii) between 40 and 50 marks.
Hint: See example 30.
4. The ages of workers in a large plant, with a mean of 50 years and standard deviation of 5 years, are assumed to be normally distributed. If 20% of the workers are below a certain age, find that age.
Hint: Given P(X < X1) = 0.20, find X1.
5. The mean and standard deviation of certain measurements computed from a large sample are 10 and 3 respectively. Use normal distribution approximation to answer the following:
(i) About what percentage of the measurements lie between 7 and 13 inclusive?
(ii) About what percentage of the measurements are greater than 16?
Hint: Apply correction for continuity.
6. There are 600 business students in the post graduate department of a university and the probability for any student to need a copy of a particular text book from the university library on any day is 0.05. How many copies of the book should be kept in the library so that the probability that none of the students, needing a copy, has to come back disappointed is greater than 0.90? (Use normal approximation to binomial.)
Hint: If X1 is the required number of copies, P(X X1) 0.90.
7. The grades on a short quiz in biology were 0, 1, 2, 3, ...... 10 points, depending upon the number of correct answers out of 10 questions. The mean grade was 6.7 with standard deviation of 1.2. Assuming the grades to be normally distributed, determine
(i) the percentage of students scoring 6 points, (ii) the maximum grade of the lowest 10% of the class.
Hint: Apply normal approximation to binomial.
8. The following rules are followed in a certain examination. "A candidate is awarded a first division if his aggregate marks are 60% or above, a second division if his aggregate marks are 45% or above but less than 60% and a third division if the aggregate marks are 30% or above but less than 45%. A candidate is declared failed if his aggregate marks are below 30% and awarded a distinction if his aggregate marks are 80% or above."

At such an examination, it is found that 10% of the candidates have failed, 5% have obtained distinction. Calculate the percentage of students who were placed in the second division. Assume that the distribution of marks is normal. The areas under the standard normal curve from 0 to z are

z:          1.28     1.64     0.41     0.47
Area : 0.4000 0.4500    0.1591   0.1808

Hint: First find parameters of the distribution on the basis of the given information.
9. For a certain normal distribution, the first moment about 10 is 40 and the fourth moment about 50 is 48. What is the mean and standard deviation of the distribution?
Hint: Use the condition b2 = 3, for a normal distribution.
10. In a test of clerical ability, a recruiting agency found that the mean and standard deviation of scores for a group of fresh candidates were 55 and 10 respectively. For another experienced group, the mean and standard deviation of scores were found to be 62 and 8 respectively. Assuming a cut-off scores of 70, (i) what percentage of the experienced group is likely to be rejected, (ii) what percentage of the fresh group is likely to be selected, (iii) what will be the likely percentage of fresh candidates in the selected group? Assume that the scores are normally distributed.
Hint: See example above.
11. 1,000 light bulbs with mean life of 120 days are installed in a new factory. Their length of life is normally distributed with standard deviation of 20 days. (i) How many bulbs will expire in less than 90 days? (ii) If it is decided to replace all the bulbs together, what interval should be allowed between replacements if not more than 10 percent bulbs should expire before replacement?
Hint: (ii) P(X X1) = 0.9.
12. The probability density function of a random variable is expressed as

(i) Identify the distribution.
(ii) Determine the mean and standard deviation of the distribution.
(iii) Write down two important properties of the distribution.
Hint: Normal distribution.
13. The weekly wages of 2,000 workers in a factory are normally distributed with a mean of Rs 200 and a variance of 400. Estimate the lowest weekly wages of the 197 highest paid workers and the highest weekly wages of the 197 lowest paid workers.
Hint: See example.
14. Among 10,000 random digits, in how many cases do we expect that the digit 3 appears at the most 950 times. (The area under the standard normal curve for z = 1.667 is 0.4525 approximately.)
Hint: m = 10000´ 0.10 and s2 = 1000*0.9.
15. Marks obtained by certain number of students are assumed to be normally distributed with mean 65 and variance 25. If three students are taken at random, what is the probability that exactly two of them will have marks over 70?
Hint: Find the probability (p) that a student gets more than 70 marks. Then find 3C2p2q.
16. The wage distribution of workers in a factory is normal with mean Rs 400 and standard deviation Rs 50. If the wages of 40 workers be less than Rs 350, what is the total number of workers in the factory? [ given ò10 f (t) dt = 0.34 , where t ~ N (0,1).]
Hint: N * Probability that wage is less than 350 = 40.
17. The probability density function of a continuous random variable X is given by
f(X) = kX(2 - X), 0 < X < 2
= 0 elsewhere.
Calculate the value of the constant k and E(X).
Hint: To find k, use the fact that total probability is unity.
18. take a screen shot of the question.
Hint: Transform X into standard normal variate z.
19. The income of a group of 10,000 persons was found to be normally distributed with mean Rs 1,750 p.m. and standard deviation Rs 50. Show that about 95% of the persons of the group had income exceeding Rs 1,668 and only 5% had income exceeding Rs 1,832.
Hint: See example 30.
20. A complex television component has 1,000 joints by a machine which is known to produce on an average one defective in forty. The components are examined and the faulty soldering corrected by hand. If the components requiring more than 35 corrections are discarded, what proportion of the components will be thrown away?
Hint: Use Poisson approximation to normal distribution.
21. The average number of units produced by a manufacturing concern per day is 355 with a standard deviation of 50. It makes a profit of Rs 150 per unit. Determine the percentage of days when its total profit per day is (i) between Rs 457.50 and Rs 645.00, (ii) greater than Rs 628.50.
Hint: Find the probabilities of producing 457.50/150 to 645/150 units etc.
22. A tyre manufacturing company wants 90% of its tyres to have a wear life of at least 40,000 kms. If the standard deviation of the wear lives is known to be 3,000 kms, find the lowest acceptable average wear life that must be achieved by the company. Assume that the wear life of tyres is normally distributed.
23. The average mileage before the scooter of a certain company needs a major overhaul is 60,000 kms with a S.D. of 10,000 kms. The manufacturer wishes to warranty these scooters, offering to make necessary overhaul free of charge if the buyer of a new scooter has a breakdown before covering certain number of kms. Assuming that the mileage, before an overhaul is required, is distributed normally, for how many kms should the manufacturer warranty so that not more than 3% of the new scooters come for free overhaul?

24. After an aeroplane has discharged its passengers, it takes crew A an average of 15 minutes (s= 4 min.) to complete its task of handling baggage and loading food and other supplies. Crew B fuels the plane and does maintenance checks, taking an average of 16 minutes (s= 2 min.) to complete its task. Assume that the two crews work independently and their times, to complete the tasks, are normally
distributed. What is the probability that both crews will complete their tasks soon enough for the plane to be ready for take off with in 20 minutes?
Hint: P(A).P(B) = P(AB).
25. An automobile company buys nuts of a specified mean diameters m. A nut is classified as defective if its diameter differs from mby more than 0.2 mm. The company requires that not more than 1% of the nuts may be defective. What should be the maximum variability that the manufacturer can allow in the production of nuts so as to satisfy the automobile company?
Hint: Find S.D.

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