Question

The mean weight of an adult is 60 kilograms with a variance of 100

If 118 adults are randomly selected, what is the probability that the sample mean would be greater than 62.2 kilograms? Round your answer to four decimal places.

Answer 1

We calculate standard deviation

n = sample size = 118

Here n is greater than 30 so according to Central limit theorem

we use normal distribution

We are asked to find

We calculate Z score for 62.2

P (Z >2.389811708)

We round above z score to 2 decimal place

P (Z >2.39)

We use Z table table to find the

and we know total are is 1

We use Z table to find

We look for row headed 2.3 and column headed 0.09

P (Z >2.39 ) = 1- 0.9914 = 0.0084

P (Z > 2.39 ) = 0.0084

Final Answer:-

0.0084

I hope this will help you :)