Question

What are the mean and standard deviation of the binomial distribution used in (a) through (c)? Interpret these values. In a survey
conducted by the Society for Human Resource Management, 68% of workers said that employers have the right to monitor their
telephone use. ( Snapshots, usatoday.com, April 18, 2006). Suppose that a random sample of 20 workers is selected, and they are
asked if employers have the right to monitor telephone use. What is the probability that:
a. 5 or less of the workers agree?
b. 10 or less of the workers agree?
c. 15 or less of the workers agree?

Answer 1

Let $X=$ the number of the workers agree: $X\sim Bin(n, p)$

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

Given $p=0.68,n=20.$

$$E(X)=np=20(0.68)=13.6$$

$$\sigma_X=\sqrt{np(1-p)}=\sqrt{20(0.68)(1-0.68)}\approx2.086$$

a. 5 or less of the workers agree

$$P(X\leq5)=P(X=0)+P(X=1)+P(X=2)+$$

$$+P(X=3)+P(X=4)+P(X=5)=$$

$$=\binom{20}{0}0.68^0(1-0.68)^{20-0}+\binom{20}{1}0.68^1(1-0.68)^{20-1}+$$

$$+\binom{20}{2}0.68^2(1-0.68)^{20-2}+\binom{20}{3}0.68^3(1-0.68)^{20-3}+$$

$$+\binom{20}{4}0.68^4(1-0.68)^{20-4}+\binom{20}{5}0.68^5(1-0.68)^{20-5}\approx$$

$$\approx0.000099$$

b. 10 or less of the workers agree

$$P(X\leq10)=P(X=0)+P(X=1)+P(X=2)+$$

$$+P(X=3)+P(X=4)+P(X=5)+P(X=6)+$$

$$+P(X=7)+P(X=8)+P(X=9)+P(X=10)\approx$$

$$\approx0.071899$$

c. 15 or less of the workers agree

$$P(X\leq15)=1-P(X>15)=1-P(X=16)-P(X=17)-$$

$$-P(X=18)-P(X=19)-P(X=20)\approx$$

$$\approx1-0.182723\approx0.817277$$