Compared to the 25 winter seasons from 1981-2006, the 35.6 inches of snowfall in

Question

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.

Explanation / Answer

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.

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