The mean preparation fee H&R Block charged retail customers in 2012 was $183 (The Wall Street Journal, March 7, 2012). Use this price as the population mean and assume the population standard deviation of preparation fees is $50. Use z-table here (link below). Round to four decimal places.a. What is the probability that the mean price for a sample of 30 H&R Block retail customers is within $8 of the population mean?b. What is the probability that the mean price for a sample of 50 H&R Block retail customers is within $8 of the population mean?c. What is the probability that the mean price for a sample of 100 H&R Block retail customers is within $8 of the population mean?d. Which, if any, of the sample sizes in parts (a), (b), and (c) would you recommend to have at least a 0.95 probability that the sample mean is within $8 of the population mean?

Answer:a)0.6192b)0.7422c)0.8904d)at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.Step-by-step explanation:Let z(p) be the z-statistic of the probability that the mean price for a sample is within the margin of error. Thenz(p)=[tex]\frac{ME*\sqrt{N}}{s }[/tex] where Me is the margin of error from the means is the standard deviation of the populationN is the sample sizea.z(p)=[tex]\frac{8*\sqrt{30}}{50 }[/tex] ≈ 0.8764by looking z-table corresponding p value is 1-0.3808=0.6192b.z(p)=[tex]\frac{8*\sqrt{50}}{50 }[/tex] ≈ 1.1314by looking z-table corresponding p value is 1-0.2578=0.7422c.z(p)=[tex]\frac{8*\sqrt{100}}{50 }[/tex] ≈ 1.6by looking z-table corresponding p value is 1-0.1096=0.8904d.Minimum required sample size for 0.95 probability is N≥[tex](\frac{z*s}{ME} )^2[/tex] where N is the sample sizez is the corresponding z-score in 95% probability (1.96)s is the standard deviation (50)ME is the margin of error (8) then N≥[tex](\frac{1.96*50}{8} )^2[/tex] ≈150.6 Thus at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.