Answer:
0.9987
Explanation:
Using Z score formula
z = (x-μ)/σ, where
x is the raw score = 52 inches
μ is the population mean = 94 inches
σ is the population standard deviation = 14 inches
z = 52 - 94/14
z =-3
Probability value from Z-Table:
P(x<52) = 0.0013499
P(x>52) = 1 - P(x<52)
P(x>52) = 1 - 0.0013499
= 0.99865
Therefore, the probability that, in a randomly selected year, the snowfall was greater than 52 inches is approximately 0.9987