) Furnace repair bills are normally distributed with a mean of 271 dollars and a standard deviation of 20 dollars. If 64 of these repair bills are randomly selected, find the probability that they have a mean cost between 271 dollars and 273 dollars.

P(271 ≤ x ≤ 273) is 0.29

Step-by-step explanation:

Since Furnace repair bills are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ/√n

Where

x = furnace repair bills

µ = mean

σ = standard deviation

n represents number of samples

From the information given,

µ = 271 dollars

σ = 20 dollars

n = 64

The probability that they have a mean cost between 271 dollars and 273 dollars is expressed as

P(271 ≤ x ≤ 273)

For x = 22,

z = (271 - 271)/(20/√64) = 0

Looking at the normal distribution table, the probability corresponding to the z score is 0.5

For x = 273

z = (273 - 271)/(20/√64) = 0.8

Looking at the normal distribution table, the probability corresponding to the z score is 0.79

Therefore,

P(271 ≤ x ≤ 273) = 0.79 - 0.5 = 0.29

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step-by-step explanation:

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