Question

Are your finances, buying habits, medical records, and phone calls really private? A real concern for many adults is that computers and the Internet are reducing privacy. A survey conducted by Peter D. Hart Research Associates for the Shell poll was re orted in USA Today. According to the survey, 41% or adults are concerned that employers are monitoring phone calls. Use the binonial distribution formula to calculate the robability or the following (a) Out of five adults, none is concemed that employers are monitoring phone calls. (Round your answer to three decimal places.) (b) Out of five aduits, all are concerned that employers are monitoring phone calls. (Round your answer to three decimal places.) (c) Out of five adults, exactly three are concerned that employers are monitoring phone calls. (Round your answer to three decimal places.)

Answer 1

Solution:

Let X be the number of adults who are concerned that employers are monitoring phone calls.

We know that $X\sim B(5,0.41)$

(a) Out of five adults, none is concerned that employers are monitoring phone calls.

Answer: We have to find here:

$P(X=0)$

Using the binomial distribution model we have:

$P(X=0)=\binom{n}{x}p^{x}(1-p)^{n-x}$

  $=\binom{5}{0}0.41^{0}(1-0.41)^{5-0}$

  $=1 \times 1 \times 0.59^5$

  $=0.071$

Therefore, the probability that out of 5 adults, none is concerned that employers are monitoring phone calls is $\boldsymbol{0.071}$

(b) Out of five adults, all are concerned that employers are monitoring phone calls.

Answer: We have to find here:

$P(X=5)$

Using the binomial distribution model we have:

$P(X=5)=\binom{n}{x}p^{x}(1-p)^{n-x}$

  $=\binom{5}{5}0.41^{5}(1-0.41)^{5-5}$

  $=1 \times 0.41^5 \times 1$

  $=0.012$

Therefore, the probability that out of 5 adults, all are concerned that employers are monitoring phone calls is $\boldsymbol{0.012}$

(c) Out of five adults, exactly three are concerned that employers are monitoring phone calls.

Answer: We have to find here:

$P(X=3)$

Using the binomial distribution model we have:

$P(X=3)=\binom{n}{x}p^{x}(1-p)^{n-x}$

  $=\binom{5}{3}0.41^{3}(1-0.41)^{5-3}$

  $=10 \times 0.41^3 \times 0.59^2$

  $=10 \times 0.068921 \times 0.3481$

  $=0.240$

Therefore, the probability that out of 5 adults, exactly three are concerned that employers are monitoring phone calls is $\boldsymbol{0.240}$