Question

The mean preparation fee H&R Block charged retail customers was $183 (The Wall Street Journal, March, 2012). Use this price as population mean deviation of preparation fee is $50 and assume the population standard a. What is the probability that the mean price for a sample of 1 within S8 of the population mean? 00 H&R Block retail customers is b. What is the probability that the mean price for a sample of 100 H&R Block retail customers is more than $190?

Answer 1

a)

std error,SE=std dev / sqrt n = 50/ sqrt 100 = 5

within 8 of population mean means

P(183-8<x<183+8)=P(175<X<191)

$Z_2 = \frac{X_2 - \mu}{SE}= \frac{ 191 - 183}{ 5} = 1.6$

$\Pr(175 \leq X \leq 191) = \Pr\left(\frac{ 175 - 183}{ 5} \le Z \le \frac{ 191 - 183}{ 5}\right)\\= \Pr(-1.6 \le Z \le 1.6)$

$= \Pr(Z \le 1.6) - \Pr(Z \le -1.6) = 0.9452 - 0.0548 = 0.8904$

b)

$\Pr(X \geq 190) = \Pr\left(Z \ge \frac{ 190 - 183}{ 5}\right) = \Pr(Z \ge 1.4)$

=1−0.9192=0.0808