Question

Suppose a large shipment of microwave ovens contained 18% defectives. If a sample of size 238 is selected, what is the probability that the sample proportion will differ from the population proportion by less than 3%? Round your answer to four decimal places.

Answer 1

P = 0.18

n = 238

$\mu_{\widehat p}$ = p = 0.18

$\sigma_{\widehat p}$ = sqrt(p(1 - p)/n)

= sqrt(0.18 * (1 - 0.18)/238)

= 0.0249

P(0.15 < $\widehat p$ < 0.21)

= P((0.15 - $\mu_{\widehat p}$ )/$\sigma_{\widehat p}$ < ($\widehat p$ - $\mu_{\widehat p}$ )/$\sigma_{\widehat p}$ < (0.21 - $\mu_{\widehat p}$ )/$\sigma_{\widehat p}$)

= P((0.15 - 0.18)/0.0249 < Z < (0.21 - 0.18)/0.0249)

= P(-1.20 < Z < 1.20)

= P(Z < 1.20) - P(Z < -1.20)

= 0.8849 - 0.1151

= 0.7698