Question

4. The average winter snowfall in Anchorage, Alaska is 74.5 inches. Suppose the standard deviation is 19.6 inches.

- Find the probability that a winter has less than 87.3 inches of snow. (Round to four decimal places.)

- Find the probability that a randomly selected winter has between 60.1 inches and 87.3 inches of snow. (Round to four decimal places.)

- The snowiest winter on record was 2011-2012, where 134.5 inches of snow fell; the least snowiest winter on record was 2014-2015, where 25.1 inches of snow fell. Which snowfall was more unusual? Explain.

Suppose you want to study the top 65% of snowiest winters.

- What percentage of winters will be excluded from your study?

- Winters with a snowfall less than what (in inches) will be excluded from your study? (Round to four decimal places.)

Answer 1

Given The average winter snowfall in Anchorage, Alaska is = 74.5 inches. and the standard deviation is = 19.6 inches.

Now since the population standard deviation is known hence Z distribution is applicable for probability calculation.

a) The probability that a winter has less than X = 87.3 inches of snow P(X< 87.3) is calculated by finding the Z score as:

Now the Probability is computed as P(Z< 0.6531), thus the probability is calculated using the excel formula for normal distribution which is =NORM.S.DIST(0.6531, TRUE), thus the probability is calculated as 0.7431.

b) The probability that a randomly selected winter has between X1 = 60.1 inches and X2 = 87.3 inches of snow, P(60.1 <X< 87.3) is calculated by finding the Z scores as:

Now the probability is computed as:

P(Z1<Z< Z2) = P(-0.7347 < Z < 0.6531) is computed using the excel formula for normal distribution which is =NORM.S.DIST(0.6531, TRUE)-NORM.S.DIST(-0.7347, TRUE), Thus the probability is computed as:

=0.7431−0.2313=0.5119

c) Given the snowiest winter on record was 2011-2012, where X1 = 134.5 inches of snow fell; the least snowiest winter on record was 2014-2015, where X2 = 25.1 inches of snow fell. Now to find the unusual value we need to find the Z score, if the Z score is less than -2 or greater tha 2 the value will be unusual.

The Z scores are:

Since both Z scores are unusual but the Z score corresponding to the value 134.5 is greater than 3 hence it is unusual.

d) Since we want to study the top 65% of snowiest winters, hence 100- 65 = 35 % of lower winters will be excluded from your study.

e) To calculate the value below which the study will be excluded we need to find the Z score corresponding to the lowest 35% or top 65% which is computed using the excel formula for normal distribution which is =NORM.S.INV(0.35), thus the Z score is computed as -0.3853.

Now using the Z score the value is calculated as:

Winters with a snowfall less than 66.9841 inches will be excluded from your study.

Note: Feel free to ask if query remains.