Question

In a survey conducted, 58% of workers said that employers have the right to monitor

their smart phone use. Suppose that a random sample of 18 workers is selected, and

they are asked if employers have the right to monitor smart phone use.

i. What is the probability that 5 or more of the workers agree?

ii. What is the probability that 4 to 12 of the workers agree?

iii. What is the probability that at least 10 of the workers NOT agree?

iv. What are the mean and standard deviation of the number of workers

agree?

Answer 1

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

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

Given $n=18, p=0.58$

i.

$$P(X\geq5)=1-P(X<5)$$

$$=1-P(X=0)-P(X=1)-P(X=2)$$

$$-P(X=3)-P(X=4)$$

$P(X=0)\approx1.65\times 10^{-7}$

$P(X=1)\approx4.11\times 10^{-6}$

$P(X=2)=0.0000482543$

$P(X=3)=0.00035539679$

$P(X=4)=0.00184044764$

$$P(X\geq5)\approx0.99775162498$$

ii.

$$P(4\leq X\leq12)=1-P(X<4)-P(X>12)$$

$$=1-0.00040792739-0.1628374012$$

$$=0.1628374012$$

iii.

$$P(X\leq8)=0.176778472$$

iv.

$$mean=E(X)=np=18(0.58)=10.44$$

$$\sigma^2=np(1-p)=18(0.58)(1-0.58)$$

$$=4.3848$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{4.3848}\approx2.0940$$