Satyagopal Mandal
Department of Mathematics
University of Kansas
Office: 624 Snow Hall  Phone: 785-864-5180
  • e-mail: mandal@math.ukans.edu
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    The Binomial Random Variable:

    The Number of Success in a certain number of trials

     

    We spoke about tossing a coin more often than you possibly wanted to hear. The experiment of tossing a coin is synonymous to any real life TRIAL that leads to either a SUCCESS or a FAILURE. In this case, we are thinking of a loaded coin because in real life trials chances of success and failures need not be equal. Such a trial is called a Bernoulli trial. Such a Bernoulli trial has the sample space S = {s, f} where s denotes the outcome "success" and f denotes the outcome "failure". Given a Bernoulli trial we can define a random variable X as follows:

    Y= 1

    If success

    Y= 0

    If failure

     

    Let us also have the notation P(success) = p. Then P(failure) = 1-p. So, P(Y=1) = p and P(Y=0) = 1-p. This random variable Y is called the Bernoulli(p) random variable.

     

    Sometime we perform one such Bernoulli trial several times and we are interested in the number of success. Suppose we perform n independent Bernoulli(p) trials. Let X be the number of success in these n trials. Then X is called a Binomial(n,p) (in short B(n,p)) random variable. So, X can assume values 0, 1, 2, 3, …, n. It may not be obvious that for r = 0, 1, 2, …, n we have

     

    P(X=r) = nCrpr(1-p) n-r.

    We do not plan to use this formula now. We plan to use some normal approximation for binomial probability.

     

    A typical situation of this type is when a pollster interviews n number of individuals to ask whether the individual would vote for a candidate or not. Then the pollster wishes to count the number of individual who would vote for the candidate. Each trial is the interviewing one individual and "success" means a YES answer.

     

    Before we proceed to use normal approximation to Binomial, let us note the following facts.

    1. The mean of X is np and variance np(1-p).

    2. We have X=X1+X2+…+ Xn where Xr is the Bernoulli(p) variable corresponding to the rth trial, for r = 1, 2, …, n. That is, Xr=1 if the rth trial is a success and Xr =0 if the rth trial is a failure. So, a B(n,p) random variable is the sum of n Bernoulli(p) random variables.

     

    Theorem: Suppose X is B(n,p) random variable. Then X = X/n is the proportion of success in these n trials. If n is large and p is not too close to 0 or 1, then the proportion X is approximately normally distributed with mean m = p and standard deviation s = (p(1-p)/n)1/2.

    This is the theorem we will be using to compute binomial probability. Please keep in mind that we get only an approximate probability and it works well enough if n is large enough.

    Remark: The notation X is not very usual. More common notation would be "bar(X)".

     

     

     

     

    Binomial Problems:

     

    Ex.1: A Lawrence bank knows that 35 percent (i.e. p=0.35) of the customers will visit the driveway counters. If 400 customers visit the bank, what is the approximate probability that more than 150 will visit the driveway counter?

     

    Solution: Suppose X is the number of customer out of 400 who would visit the driveway counter. The X is B(400, 0.35)-variable. The mean of the proportion X = X/400 is m = p = 0.35

    and so the standard deviation

    s = (p(1-p)/n)1/2 = (0.35 x 0.65 /400) 1/2 = 0.024.

    So, we have

    P(X > 150) = P(X/n > 150/400) =

    P(X > 0.375) =

    P((X-m)/s > (0.375-m)/s) =

    P(Z > (0.375 - 0.35)/0.024) =

    P(Z > 1) =

    1 - P(Z < 1) = 1- 0.8413 = 0.1587.

     

    Ex.2: It is known that the probability that a household owns a pressure cooker is p = 0.1. If 190 households are interviewed, find the approximate probability that

    1. more than 25 households own a food processor;

    2. less than 30 households own a food processor.

     

    Solution 1) Here n = 190 and p = 0.1. Suppose X is the number of households among these 190 who own a pressure cooker. We are asked to find P(X > 25).

    The mean of the proportion X = X/190 is m = p = 0.1

    and so the standard deviation

    s = (p(1-p)/n)1/2 = (0.1 x 0.9/190) 1/2 = 0.022.

    So, we have

    P(X > 25) = P(X/n >25/190) =

    P(X > 0.13) =

    P((X-m)/s > (0.13-m)/s) =

    P(Z > (0.13-0.1)/0.022) =

    P(Z > 1.4) =

    1 - P(Z < 1.4) = 1- 0.9192 = 0.0808.

     

    Solution 2) Here we are asked to compute P(X < 30).

    We have

    P(X < 30) = P(X/n < 30/190) =

    P(X < 0.16) =

    P((X-m)/s < (0.16-m)/s) =

    P(Z < (0.16-0.1)/0.022) =

    P(Z < 2.7) =0.9965.

     

    Ex.3: The campaign committee of a candidate claims that sixty percent (p = 0.6) of the voters are in favor of the candidate. You interview 150 voters. Assuming the campaign committee's claim, find what is the approximate probability that less than 87 will favor the candidate?

     

    Solution: Here n = 150 and p = 0.6. Suppose X is the number of voters among these 150 who are in favor of this candidate. Then X is a B(n,p) = B(150,0.6) -variable.

    The mean of the proportion X = X/150 is m = p = 0.6

    and so the standard deviation

    s = (p(1-p)/n)1/2 = (0.6 x 0.4/150) 1/2 = 0.04.

    So, we have

    P(X < 87) = P(X/n < 87/150) =

    P(X < 0.58) =

    P((X-m)/s < (0.58-m)/s) =

    P(Z < (0.58 - 0.6)/0.04) =

    P(Z < -0.5) = 0.3085.

     

    Ex. 4: A technique is used to fertilize eggs in a fertility clinic laboratory. It is known that the probability that an egg will be fertilized by this technique is p = 0.1. If 500 eggs are treated what is the probability that at least 60 eggs will be fertilized?

     

    Solution: Here n = 500 and p = 0.1. Suppose X is the number of eggs out of these 500 eggs that got fertilized. Then X is B(n,p) = B(500,0.1)-variable. We are asked to find P(X > 60).

    The mean of the proportion X = X/500 is m = p = 0.1

    and so the standard deviation

    s = (p(1-p)/n)1/2 = (0.1 x 0.9/500) 1/2 = 0.013.

    So, we have

    P(X > 60) = P(X/n > 60/500) =

    P(X > 0.12) =

    P((X-m)/s > (0.12-m)/s) =

    P(Z > (0.12 - 0.1)/0.013) =

    P(Z > 1.5) =1 - P(Z < 1.5) = 1- 0.9332 = 0.0668.

     

     

    Ex.5: The probability that a computer chip produced in a factory is defective is p = 0.2. If you have a sample of 60 chips, what is the probability that number of defective will be less than 20?

     

    Solution: Here n= 60 and p = 0.2. Suppose X is the number of defective items in these 60 chips. We asked to compute P(X < 20).

    The mean of the proportion X = X/60 is m = p = 0.2

    and so the standard deviation

    s = (p(1-p)/n)1/2 = (0.2 x 0.8/60) 1/2 = 0.051.

    So, we have

    P(X < 20) = P(X/n < 20/60) =

    P(X < 0.33) =

    P((X-m)/s < (0.33-m)/s) =

    P(Z < (0.33 - 0.2)/0.051) =

    P(Z < 2.5) = 0.9938.

     

    Ex.6: The probability that a lamp produced by a machine is defective is p = 0.2. Suppose a quality control inspector takes a sample of 120 lamps. What is the probability that more than 30 tubes will be defective?

     

    Ex.7: Suppose the probability that a student has access to Internet is p = 0.8. Suppose you interview 160 students. What is the probability that less than 120 students will have access to Internet.

     

    Ex.8: Suppose that the probability that a person favors medical use of marijuana is p = 0.6. If 780 individuals are interviewed what is the probability that more less than 450 will be in favor.

     

    Ex.9: Suppose that the probability that a middle income family invests in the stock market is p = 0.8. If we interview 880 middle income families what is the probability that more than 700 would have invested in stock market?

     

    Ex.10: Suppose that the probability an insurance company knows from experience that the probability that a life insurance-policy holder will survive another 10 years is p = 0.9. The company has 2280 policy. What is the probability that more than 2025 will survive another 10 years.