Answer:
The standard deviation is 4.83 inches.
Explanation:
We are given that the average yearly snowfall in Chillyville is Normally distributed with a mean of 55 inches.
The snowfall in Chillyville exceeds 60 inches in 15% of the years, we have to find the standard deviation.
Let X = <u><em>the average yearly snowfall in Chillyville</em></u>.
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = mean amount of rainfall = 55 inches
= standard deviation
Now, it is stated that the snowfall in Chillyville exceeds 60 inches in 15% of the years, that means;
P(X > 60 inches) = 0.15
P( >
) = 0.15
P(Z > ) = 0.15
In the z table, the critical value of z that represents the top 15% of the area is given as 1.0364, that means;
= 4.83 inches
Hence, the standard deviation is 4.83 inches.
