Question

Suppose that the monthly incomes of all employees in Firm A are normally distributed with a mean of $1784.85 and a standard deviation of $199.23. If a random sample of 121 employees is selected from Firm A, what is the probability that the mean of the employees in the sample will be greater than $1809.572632 (approximately)? 

Answer 1

$\bar{x}=1784.85 \\ σ = 199.23 \\ n = 121 \\ P(X>1809.572632) = P(\frac{x-μ}{σ/ \sqrt{n}} > \frac{1809.572632 -1784.85}{199.23/ \sqrt{121}}) \\ = P(Z>1.365) \\ = 1 -P(Z<1.365)$

From Z-table

$= 1 -0.9138 \\ = 0.0862$

Answer: 0.0862