The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 8.3 cm. a. Find the probability that an individual distance is greater than 214.30 cm. b. Find the probability that the mean for 20 randomly selected distances is greater than 202.80 cm. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30? a. The probability is. 

a. Find the probability that an individual distance is greater than 214.30 cm We find for the value of z score using the formula: z = (x – u) / s z = (214.30 – 205) / 8.3 z = 1.12 Since we are looking for x > 214.30 cm, we use the right tailed test to find for P at z = 1.12 from the tables: P = 0.1314   b. Find the probability that the mean for 20 randomly selected distances is greater than 202.80 cm We find for the value of z score using the formula: z = (x – u) / s z = (202.80 – 205) / 8.3 z = -0.265 Since we are looking for x > 202.80 cm, we use the right tailed test to find for P at z = -0.265 from the tables: P = 0.6045   c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30? I believe this is because we are given the population standard deviation sigma rather than the sample standard deviation. So we can use the z test.