Suppose the weekly amount of money spent on maintenance and repairs by a company is NOT approximately normally distributed. However, its mean and standard deviation are known to be $400 and $20, respectively. If $450 is budgeted for next week, what can you say about the probability that the actual costs will exceed the budgeted amount

Probability = 0.00621

Step-by-step explanation:

We are given;

Population mean; μ = 400

Population standard deviation; σ = 20

Raw score; x¯ = 450

the probability that the actual costs will exceed the budgeted amount will be the p-value.

Thus, let's first find the z-score after which we will find the p-value.

z = (x¯ - μ)/σ

z = (450 - 400)/20

z = 50/20

z = 2.5

From online p-value from z-score calculator attached, using z = 2.5; one tailed hypothesis; significance level of 0.05, we have;

P-value = 0.00621

5050

hope this !