Question

Random digit dialing When a polling company calls a telephone number at random, there is only a 9% chance that the call reaches a live person and the survey is successfully completed. 10 Suppose the random digit dialing machine makes 15 calls. Let X= the number of calls that result in a completed survey.

a. Find the probability that more than 12 calls are not completed.

b. Calculate and interpret $\mu {\mathrm{X}}_X$.

c. Calculate and interpret $\sigma \mathrm{X}{\sigma }_X$.

Short answer

(a) the required probability is $0.8531$.

(b)The computed value of mean shows that on an average out of 15 calls , $1.35$will be completed.

(c)The total number of calls that get completed will vary by $1.1084$calls in comparison of mean of $1.35$calls.

Step-by-step solution

Part (a) Step 1: Given Information

Probability of success $\left(p\right)=9\%=0.09$

Number of trials $\left(n\right)=15$

The mean and standard deviation are calculated using the following formula:

$Mean\left(\mu \right)=n\times p$

Standard deviation $\left(\sigma \right)=\sqrt{n\times p\times (1-p)}$

Part (a) Step 2: Simplification

Assume X to be the random number of calls that aid in the completion of the survey, which follows the binomial distribution.

If more than 12 calls go unanswered, that suggests fewer than 3 calls were made.

The following formula can be used to compute the probability:

$$\begin{gathered} P(X<3)=P(X=0)+P(X=1)+P(X=2) \\ =\sum _{r=0}^2{}^{15}C_r\times (0.09{)}^r\times (1-0.09{)}^{15-r} \\ =0.8531 \end{gathered}$$

Thus, the required probability is $0.8531$.

Part (b) Step 1:Given information

Given:

Probability of success $\left(p\right)=9\%=0.09$

Number of trials $\left(n\right)=15$

Part (b) Step 2: Simplification

The mean of X can be computed using the formula:

$\begin{array}{rcl}{\mu }_X& =& n\times p\\ & =& 15(0.09)\\ & =& 1.35\end{array}$

Therefore, the value of mean is 1.35

Part (c) step 1: Given information

Given:

The Probability of success$\left(p\right)=9\%=0.09$

The number of trials $\left(n\right)=15$

Part (c) Step 2: Simplification

X's standard deviation can be determined as follows:

$\begin{array}{rcl}{\sigma }_X& =& \sqrt{n\times p\times (1-p)}\\ & =& \sqrt{15(0.09)(1-0.09)}\\ & =& 1.1084\end{array}$