144k views
5 votes
) 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.

User Johnny Wu
by
8.7k points

1 Answer

3 votes

Answer: P(271 ≤ x ≤ 273) is 0.29

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

User MSIL
by
8.7k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories