Compared to the 25 winter seasons from 19812006 the 356 inch

Compared to the 25 winter seasons from 1981-2006, the 35.6 inches of snowfall in Chicago during the 2006-2007 season was snowier than the 53.98% of the them and the 60.3 inches in the 2007-2008 season was more snowfall than 99.65% of them. A) Calculate the amount of snowfall needed in the 2008-2009 season to maintain the average established from 1981-2006 seasons. B) Chicago got 52.7 inches of snowfall in the 2008-2009 winter season. Calculate the percentile to accurately complete the statement: The 2008-2009 winter saw more snowfall that ______% of winters from 1981-2006. C) Find the average seasonal snowfall from 1981 to 2009 to the nearest tenth of an inch.

Solution

We assume that the snow falls are normally distributed.We have the probability and we need to find the inverse z score values

P (Z < a) = 0.5398

a = 0.1

Thus.

35.6 = mu + 0.1(sd)

Now,

P(Z<b) = 0.9965

b = 2.697

60.3 = mu + 2.697 (sd)

Solving for mean and sd, we get,

mu = 34.6489 ~ 34.7

sd = 9.5109 ~ 9.5

A)

Mean of 25 seasons = 34.6489

2006-07 : 35.6

2007-08 : 60.3

2008-09 : x

Thus, (25 * 34.7 + 35.6 + 60.3 + x)/28 = 34.7

x = 8.2 inches

b)

Z score = (52.7 - 34.7 )/9.5

= 1.894

P (Z < 1.894) = 0.9708

Thus, 97.08% of winters

C)

Average of all the seasons so far = (25*34.7 + 35.6 + 60.3 + 52.7) /28

= 36.289

~ 36.3 inches

Hope this helps. Ask if you have doubts.