Suppose personal daily water usage in New York City is normally distributed, wit

Question

Suppose personal daily water usage in New York City is normally distributed, with a mean of 18 litres and a standard deviation of 6 litres.

a) What percentage of the population uses between 10 and 20 litres?

b) What is the probability that a person uses less than 10 litres?

c) Suppose the local government wants to give a tax rebate to the 20% of the population that use the least amount of water. What should the government use as the maximum water limit for a person to qualify for a tax rebate?

d) Suppose the governments’ proposed tax rebate causes a shift in the average water use from 18 litres to 14 litres per person per day, but causes no shift in the standard deviation. What limit should be set on water use if 20% of the population is to receive a tax rebate?

Explanation / Answer

a. Assume r.v denote the personal daily water usage in New York City.

P(10<X<20)=P[(10-18)/6<Z<(20-18)/6] [use Z=(X-mu)/sigma, where, X is raw score, mu is population mean, and sigm ai spopulation standard deviation]

=P[-1.33<Z<0.33] (the two Z scores are of opposite signs, therefore, find areas between respectiev Z scores and the mean and add them)

=0.4082+0.1293=0.5375

b. P(X<10)=P(Z<(10-18)/6]=P(Z<-1.33)=0.0918

c. Assume Y denote the limit to be placed. From information given, P(X<=x)=0.20, find Z score corresponding to area closest to 0.20. The closest value is 0.2005. The corresponding Z score is -0.84. Then substitute the values in the Z score formula to obtain the raw score.

-0.84=(Y-18)/6, Y=12.96

d. Assume Z denote the limit to be placed. Replace 18 with 14.

-0.84=(Z-14)/6, Z=8.96

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