Question

when a polling company calls a telephone number at random, there is only a 9% probability that the call reaches a live person and the survey is successfully completed. suppose the random digit dialing machine makes 15 calls. let x = the number of calls that result in a completed survey.
(a) calculate the mean of x.
(round to 2 decimal places.)
interpret the mean of x.
if 15 phone calls are made by the random dialing machine, we expect about surveys will be completed, on average.
(b) calculate the standard deviation of x.
(round to 2 decimal places.)
interpret the standard deviation of x.
if 15 phone calls are made by the random dialing machine, the number of completed surveys would typically vary from by about completed surveys.

Explanation:

Step1: Identify the distribution

This is a binomial distribution problem where \( n = 15 \) (number of trials) and \( p = 0.09 \) (probability of success). For a binomial distribution, the mean \( \mu = np \) and the standard deviation \( \sigma=\sqrt{np(1 - p)} \).

Step2: Calculate the mean (part a)

Using the formula for the mean of a binomial distribution: \( \mu=np \). Substitute \( n = 15 \) and \( p=0.09 \):
\( \mu=15\times0.09 = 1.35 \)

Step3: Calculate the standard deviation (part b)

First, find \( 1 - p=1 - 0.09 = 0.91 \). Then use the standard deviation formula for binomial distribution \( \sigma=\sqrt{np(1 - p)} \). Substitute \( n = 15 \), \( p = 0.09 \) and \( 1 - p=0.91 \):
\( \sigma=\sqrt{15\times0.09\times0.91}=\sqrt{15\times0.0819}=\sqrt{1.2285}\approx1.11 \)

Answer:

(a) The mean of \( X \) is \( 1.35 \).
(b) The standard deviation of \( X \) is approximately \( 1.11 \).