(10+ 10+ 10 Points] Phoenix water is provided to approximately 1.4 million peopl
ID: 2931310 • Letter: #
Question
(10+ 10+ 10 Points] Phoenix water is provided to approximately 1.4 million people who are served through more than 362,000 accounts (http://phoenix.gov/WATER/wtrfacts.html). All accounts are metered and billed monthly. The probability that an account has an error in a month i be assumed to be independent. (a) What are the mean and standard deviation of the number of account errors each month? (b) Find the probability of fewer than 345 errors in a month. (c) Find a value so that the probability that the number of errors exceeds this value is 0.10. 3.Explanation / Answer
(a)
Let Number of accounts be n and Probability of account in error be p.
Then n = 362000 and p = 0.001
Assuming that the number of accounts error follow binomial distribution,
Mean number of account errors = n * p = 362000 * 0.001 = 362
Variance of number of account errors = n p (1-p)
= 362000 * 0.001 * (1 - 0.001) = 361.638
Standard deviation of number of account errors = sqrt(361.638) = 19.02
(b)
As, np > 5 and n(1-p) > 5, we can approximate the binomial distribution of number of accounts error as a normal distribution with mean 362 and standard deviation of 19.02
Probability of fewer than 345 errors in a month = P(X < 345)
= P(Z < (345-362)/ 19.02)
= P(Z < -0.8938)
= 0.1857 (Using Z table)
Probability of fewer than 345 errors in a month is 0.1857
(c)
Let x be the value that the probability that the number of errors exceeds this value is 0.10. So,
P(Z > z) = 0.1
Using Z table, we get z = 1.28
Then,
1.28 = (x - 362) / 19.02
=> x - 362 = 1.28 * 19.02 = 24.35
=> x = 362 + 24.35 = 386.35
So, 386.35 is the value such that the probability that the number of errors exceeds this value is 0.10
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