Question

The amount of snow fall falling in a certain mountain range is normally distributed with a mean of 74 inches, and a standard deviation of 12 inches. what is the probability that the mean annual snowfall during 36 randomly picked years will exceed 76.8 inches?

Answer 1

Given that the sample mean is μ=74 and sample standard deviation σₓ=$\frac{σ}{ \sqrt{n}} =\frac{12}{\sqrt{36}} =2$.

Thge Z- score is $Z=\frac{76.8-74}{2} =1.4$.

Now we use the Z-score to calculate the required probability.

Refer to standard normal distribution table.

The required probability is

$P(X>76.8) =P(Z>1.4)=1-P(Z$