Question

I need help with parts b) and c). Please eplain how you determined P(x overscore <72.3) and P(x overscore >or equal to 70.9)

A simple random sample of size n 42 is obtained from a population with - 69 and a = 19. (a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample mean? Assuming that this condition is true, describe the sampling distribution of x. (b) Assuming the normal model can be used, determine P(x < 72.3) (c) Assuming the normal model can be used, determine P(x70.9) A. The population must be normally distributed. B. Since the sample size is large enough, the population distribution does not need to be normal C. The population must be normally distributed and the sample size must be large. D. The sampling distribution must be assumed to be normal Assuming the normal model can be used, describe the sampling distribution x. Choose the correct answer below. 69 and = 19 O A Approximately normal, with 42 B. Approximately normal, with = 69 and o% 19 19 C. Approximately normal, with 69 and o- 42 'x (b) P(x<72.3) |(Round to four decimal places as needed.)

Answer 1

Solution:

Given in the question

Mean = 69, and population standard deviation = 19

Solution(a)

As sample size is greater than 30 and population standard devaition is also known so population distribution need not to be normally distributed.

Assuming population is normal than mean = 69 and standard deviation = 19/sqrt(42)

Solution(b)

We need to calculate P(Xbar<72.3)

Z = (72.3-69)/19/sqrt(42) = 1.13

From Z table we found p-value

P(Xbar<72.3) = 0.8708

Solution(c)

P(Xbar>=70.9) = 1- P(Xbar<70.9)

Z = (70.9-69)/19/sqrt(42) = 0.648

From z table we found p- value

P(Xbar>=70.9) = 1- 0.7422 = 0.2578