Question

1. Suppose that personal daily water usage in California is normally distributed with a mean of 17 gallons and a standard deviation of 6 gallons.

(a) What proportion of California’s population uses between 10 and 20 gallons daily?

(b) The governor of California wants to give a tax rebate to the 20% of the population that uses the least amount of water. What should the governor use as the maximum daily water usage for a person to qualify for the tax rebate?

(c) The governor randomly samples 100 people from the California population. What is the probability that the average usage among these 100 people is more than 18 gallons/day?

Answer 1

This is a normal distribution question with

a)

P(10.0 < x < 20.0)=?

This implies that

P(10.0 < x < 20.0) = P(-1.1667 < z < 0.5) = P(Z < 0.5) - P(Z < -1.1667)

P(10.0 < x < 20.0) = 0.6914624612740131 - 0.12166577141578738

b) Given in the question

P(X < x) = 0.2

This implies that

P(Z < -0.8416212335729142) = 0.2

With the help of formula for z, we can say that

c) Sample size (n) = 100

Since we know that

P(x > 18.0)=?

The z-score at x = 18.0 is,

z = 1.6667

This implies that

P(x > 18.0) = P(z > 1.6667) = 1 - P(z < 1.6667) = 1 - 0.9522129635397043

PS: you have to refer z score table to find the final probabilities.

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