Question

STA2023(IAC)SATURDAY-FALL-2019 Homework: HOMEWORK #4 Score: 0 of 1 pt 6.4.6-T 10 of 35 (22 complete The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm and a standard deviation of 8.3 cm a. Find the probability that an individual distance is greater than 210.90 cm b. Find the probability that the mean for 15 randomly selected distances is greater than 196.20 cm. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 307 a. The probability is (Round to four decimal places as neede ve More Enter your answer in the answer box and then click Check Answer. 2 parts remaining

Answer 1

Ans: Mean (M) = 197.5m Std dev (0) = 8.3 eu I Let x be an overlad reach distance of adett XNN(197.5, 8.3) 1a P(xs 210.90) = 8(x-M 210.9 - 197.5) - Plz> 1.62) = 1-8(261.62) = 1-0,94738 =10.0526) 8.3

p(x > 196.2) - Plu s 1.6.2 - 197.5) 8.31015 P (2>-0,61) 18 (22-0.61) = 1-0.27693 (0.9291 TO According to central lomit theorem If population is normal, then Samplong distribution of the sample means woll be approximately nemal een for Samples of size less than 30