The area of a rectangle is 144 square centimeters. The width is 9 centimeters. Which of the following statements is true? Select
2 answers:
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Given that the population mean, 
and the population standard deviation, 
Part A:
<span>The probability that the mean price for a sample of 30 h&r block retail customers is within $8 of the population mean is evaluated as follows:
![P(183-8\leq\bar{x}\leq183+8)=P(175\leq\bar{x}\leq191) \\ \\ =P(\bbar{x}\leq191)-P(\bbar{x}\leq175)=P\left(z\leq \frac{191-183}{ \frac{50}{ \sqrt{30} } } \right)-P\left(z\leq \frac{175-183}{ \frac{50}{ \sqrt{30} } } \right) \\ \\ =P(z\leq0.8764)-P(z\leq-0.8764) \\ \\ =P(z\leq0.8764)-[1-P(z\leq0.8764)] \\ \\ =2P(z\leq0.8764)-1=2(0.80958)-1 \\ \\ =1.61916-1=0.61916\approx62\%](https://tex.z-dn.net/?f=P%28183-8%5Cleq%5Cbar%7Bx%7D%5Cleq183%2B8%29%3DP%28175%5Cleq%5Cbar%7Bx%7D%5Cleq191%29%20%5C%5C%20%20%5C%5C%20%3DP%28%5Cbbar%7Bx%7D%5Cleq191%29-P%28%5Cbbar%7Bx%7D%5Cleq175%29%3DP%5Cleft%28z%5Cleq%20%5Cfrac%7B191-183%7D%7B%20%5Cfrac%7B50%7D%7B%20%5Csqrt%7B30%7D%20%7D%20%7D%20%5Cright%29-P%5Cleft%28z%5Cleq%20%5Cfrac%7B175-183%7D%7B%20%5Cfrac%7B50%7D%7B%20%5Csqrt%7B30%7D%20%7D%20%7D%20%5Cright%29%20%5C%5C%20%20%5C%5C%20%3DP%28z%5Cleq0.8764%29-P%28z%5Cleq-0.8764%29%20%5C%5C%20%20%5C%5C%20%3DP%28z%5Cleq0.8764%29-%5B1-P%28z%5Cleq0.8764%29%5D%20%5C%5C%20%20%5C%5C%20%3D2P%28z%5Cleq0.8764%29-1%3D2%280.80958%29-1%20%5C%5C%20%20%5C%5C%20%3D1.61916-1%3D0.61916%5Capprox62%5C%25)
Part B:
</span><span>The probability that the mean price for a sample of 50 h&r block retail customers is within $8 of the population mean is evaluated as follows:
</span>![P(183-8\leq\bar{x}\leq183+8)=P(175\leq\bar{x}\leq191) \\ \\ =P(\bbar{x}\leq191)-P(\bbar{x}\leq175)=P\left(z\leq \frac{191-183}{ \frac{50}{ \sqrt{50} } } \right)-P\left(z\leq \frac{175-183}{ \frac{50}{ \sqrt{50} } } \right) \\ \\ =P(z\leq1.131)-P(z\leq-1.131) \\ \\ =P(z\leq1.131)-[1-P(z\leq1.131)] \\ \\ =2P(z\leq1.131)-1=2(0.87105)-1 \\ \\ =1.7421-1=0.7421\approx74\%](https://tex.z-dn.net/?f=P%28183-8%5Cleq%5Cbar%7Bx%7D%5Cleq183%2B8%29%3DP%28175%5Cleq%5Cbar%7Bx%7D%5Cleq191%29%20%5C%5C%20%5C%5C%20%3DP%28%5Cbbar%7Bx%7D%5Cleq191%29-P%28%5Cbbar%7Bx%7D%5Cleq175%29%3DP%5Cleft%28z%5Cleq%20%5Cfrac%7B191-183%7D%7B%20%5Cfrac%7B50%7D%7B%20%5Csqrt%7B50%7D%20%7D%20%7D%20%5Cright%29-P%5Cleft%28z%5Cleq%20%5Cfrac%7B175-183%7D%7B%20%5Cfrac%7B50%7D%7B%20%5Csqrt%7B50%7D%20%7D%20%7D%20%5Cright%29%20%5C%5C%20%5C%5C%20%3DP%28z%5Cleq1.131%29-P%28z%5Cleq-1.131%29%20%5C%5C%20%5C%5C%20%3DP%28z%5Cleq1.131%29-%5B1-P%28z%5Cleq1.131%29%5D%20%5C%5C%20%5C%5C%20%3D2P%28z%5Cleq1.131%29-1%3D2%280.87105%29-1%20%5C%5C%20%5C%5C%20%3D1.7421-1%3D0.7421%5Capprox74%5C%25)
Part C:
<span>The probability that the mean price for a sample of 50 h&r block retail customers is within $8 of the population mean is evaluated as follows:
![P(183-8\leq\bar{x}\leq183+8)=P(175\leq\bar{x}\leq191) \\ \\ =P(\bbar{x}\leq191)-P(\bbar{x}\leq175)=P\left(z\leq \frac{191-183}{ \frac{50}{ \sqrt{100} } } \right)-P\left(z\leq \frac{175-183}{ \frac{50}{ \sqrt{100} } } \right) \\ \\ =P(z\leq1.6)-P(z\leq-1.6) \\ \\ =P(z\leq1.6)-[1-P(z\leq1.6)] \\ \\ =2P(z\leq1.6)-1=2(0.9452)-1 \\ \\ =1.8904-1=0.8904\approx89\%](https://tex.z-dn.net/?f=P%28183-8%5Cleq%5Cbar%7Bx%7D%5Cleq183%2B8%29%3DP%28175%5Cleq%5Cbar%7Bx%7D%5Cleq191%29%20%5C%5C%20%5C%5C%20%3DP%28%5Cbbar%7Bx%7D%5Cleq191%29-P%28%5Cbbar%7Bx%7D%5Cleq175%29%3DP%5Cleft%28z%5Cleq%20%5Cfrac%7B191-183%7D%7B%20%5Cfrac%7B50%7D%7B%20%5Csqrt%7B100%7D%20%7D%20%7D%20%5Cright%29-P%5Cleft%28z%5Cleq%20%5Cfrac%7B175-183%7D%7B%20%5Cfrac%7B50%7D%7B%20%5Csqrt%7B100%7D%20%7D%20%7D%20%5Cright%29%20%5C%5C%20%5C%5C%20%3DP%28z%5Cleq1.6%29-P%28z%5Cleq-1.6%29%20%5C%5C%20%5C%5C%20%3DP%28z%5Cleq1.6%29-%5B1-P%28z%5Cleq1.6%29%5D%20%5C%5C%20%5C%5C%20%3D2P%28z%5Cleq1.6%29-1%3D2%280.9452%29-1%20%5C%5C%20%5C%5C%20%3D1.8904-1%3D0.8904%5Capprox89%5C%25)
</span>
Part A:
<span>The probability that the mean price for a sample of 30 h&r block retail customers is within $8 of the population mean is evaluated as follows:
Part B:
</span><span>The probability that the mean price for a sample of 50 h&r block retail customers is within $8 of the population mean is evaluated as follows:
</span>
Part C:
<span>The probability that the mean price for a sample of 50 h&r block retail customers is within $8 of the population mean is evaluated as follows: