e 9-month salaries at a daycare centernormally distributed with a mosnof$20,00 D

Question

e 9-month salaries at a daycare centernormally distributed with a mosnof$20,00 Dnda standard deviation of $5,000 d. What is the probability that an employee will have a salary between $18,000 and $22,0001 What is the probability that an employee will have a salary more tham $18,000? e.What is the probability that an employee will have a salary less thaen $22,000? In order to estimate the average electric usage per month, a sample of 100 houses was selecte d the electric usage determined Assume a population standard deviation of 200 kilowatt hours. Determine the standard ero mean.

Explanation / Answer

Given: µ = $20,000, = $5000

To find the probability, we need to find the Z scores first.

Z = (X - µ)/ [/n]. Since n = 1, Z = (X - µ)/

a) For P (18000 < X < 22000) = P(X < 22000) – P(X < 18000)

For P( X < 22000)

Z = (22000 – 20000)/5000 = 0.4

The probability for P(X < 22000) from the normal distribution tables is = 0.6554

For P( X < 18000)

Z = (18000 – 20000)/5000 = -0.4

The probability for P(X < 18000) from the normal distribution tables is = 0.3446

Therefore the required probability is 0.6554 – 0.3446 = 0.3108

b) For P (X > 18000) = 1 - P (X < 18000), as the normal tables give us the left tailed probability only.

For P( X < 18000) = 0.3446

Therefore the required probability = 1 – 0.3446 = 0.6554

c) For P( X < 22000), we have already found the same in (a)

The required probability from the normal distribution tables is = 0.6554

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