Furnace repair bills are normally distributed with a mean of 263 dollars and a standard deviation of 25 dollars. If 100 of these repair bills are randomly selected, find the probability that they have a mean cost between 263 dollars and 265 dollars.

0.28814

Step-by-step explanation:

Given that (m) = 263

Standard deviation (s) = 25

Sample size (n) = 100

find the probability that they have a mean cost between 263 dollars and 265 dollars

Using the z formula :

Zscore = (x - m) / (s/√n)

For x = 263

Zscore = (263 - 263) / (25/√100) = 0

P(z < 0) = 0.5 ( Z probability calculator)

For x = 265

Zscore = (265 - 263) / (25/√100) = 0

Zscore = 2 / (25/10)

Zscore = 2 / 2.5

Zscore = 0.8

P(z < 0.8) = 0.78814 ( Z probability calculator)

(0.78814 - 0.5) = 0.28814