Question

Suppose that a random sample of size 100 is to be selected from a population with mean 50 and standard deviation 8. What is the probability that a sample mean will be greater than 50.8? Round your answer to three decimal places.

Answer 1

Answer:

0.159

Step-by-step explanation:

Given that:

Sample size (n) = 100

Population mean(pm) = 50

Standard deviation (s) = 8

Probability that Sample mean (m) will be greater Than 50.8

Using the relation :

(sample mean - population mean) / (standard deviation /sqrt(n))

P(m > 50.8)

Z = (50.8 - 50) / (8/ sqrt(100))

Z = 0.8 / (8/ 10)

Z = 0.8 / 0.8

Z = 1

P(Z > 1) = 0.15866 ( Z probability calculator)

Hence,

P(Z > 1) = 0.159 ( 3 decimal places)