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) = [tex]\sqrt{\frac{p(1-p)}{n} }[/tex] = [tex]\sqrt{\frac{0.53(1-0.53)}{250} }[/tex] = 0.0316
For x = 120, sample proportion(p) = [tex]\frac{x}{n}[/tex] = [tex]\frac{120}{250}[/tex] = 0.48
So, likelihood that fewer than 120 have a PC
= P(x < 120)
= P( p^ < 0.48 )
= P(z < [tex]\frac{0.48-0.53}{0.0316}[/tex]) (z=[tex]\frac{p^-p}{SE}[/tex])
= P(z < -1.58)
= 0.0571 ( From normal table )
