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?

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) =   = = 0.0316

For x = 120, sample proportion(p) = = = 0.48

So, likelihood that fewer than 120 have a PC

= P(x < 120)

= P(  p^  < 0.48 )

= P(z < )      (z=)  

= P(z < -1.58)

= 0.0571      ( From normal table )

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