Question

Fifty-three percent of U.S households have a personal computer. In a random sample of 250 households, what is the probability that fewer than 120 have a PC?

Answer 1

Answer:

The correct Answer is 0.0571

Explanation:

53% of U.S. households have a PCs.

So, P(Having personal computer) = p = 0.53

Sample size(n) = 250

np(1-p) = 250 * 0.53 * (1 - 0.53) = 62.275 > 10

So, we can just estimate binomial distribution to normal distribution

Mean of proportion(p) = 0.53

Standard error of proportion(SE) =  $\sqrt{\frac{p(1-p)}{n} }$ = $\sqrt{\frac{0.53(1-0.53)}{250} }$ = 0.0316

For x = 120, sample proportion(p) = $\frac{x}{n}$ = $\frac{120}{250}$ = 0.48

So, likelihood that fewer than 120 have a PC

= P(x < 120)

= P(  p^​  < 0.48 )

= P(z < $\frac{0.48-0.53}{0.0316}$​)      (z=$\frac{p^-p}{SE}$​)  

= P(z < -1.58)

= 0.0571      ( From normal table )