Statistics section 1.3

Question

Suppose that water usage in American showers are normally distributed, with an average shower using 16.7 gallons, and a standard deviation of 2.9 gallons. Estimate the percentage of showers that used

a) .... % of showers used between 13.8 and 19.6 gallons

b) ....% of showers use more than 25.4 gallons

c) ....% of showers use less than 13.8 gallons

d) according to the 2 deviation rule data values that are more than 2 standard dev away from the mean of the distribution are considered to be unusually high or unusually low. According to this rule it would be unusual for an American shower to last more than .... gallons?

Answer 1

Given μ = 16.7 σ = 2.9

Answers to be calculated using a normalized cumulative distribution function. (a standard normal distribution )

Similar calculations are directly calculable using cumulative distribution functions with x, μ, and σ as inputs.

a) since 13.8 and 19.6 both deviate from the mean (16.7) by the same amount (1 σ or 2.9) we only have to calculate how much area of the SND is between 1 σ and 0 and double it.

P(0 ≤ z ≤ 1) = .34134 answer = .68568

b) Z score for 25.4 gal = (25.4 - 16.7)/2.9 or 3.0 ; P (3 ≤ z) = .00135

c) Z score for 13.8 gal (from above) is -1 P(z ≤ -1) = .15866

d) opps. forgot to answer this. Here it is. Using the 2 sigma rule as being excessive, 2 sigma (2x2.9) more than the average/mean = 16.7 + 5.8 or 22.5 gal. So we would expect a shower using more than 22.5 gal to be unusually long.

Answer 2

A) 68%

B) 0.1%

C) 15.8%

D) 25.4 gal