Question

Furnace repair bills are normally distributed with a mean of 269 dollars and a standard deviation of 20 dollars. a. If 64 of these repair bills are randomly selected, calculate that standard error of the mean for the sampling distribution of 64 repair bills. b. Using the standard error from part a, find the probability that they have a mean cost between 269 dollars and 271 dollars.

Answer 1

Given:

μ = 269

σ = 20

n = 64

a) Standard Error = σ / √n

= 20 / √64

= 2.5

                Standard Error = 2.5

b) SInce the data is normally distributed, we use the formula;

Z = ( x – μ ) / σ

               For x = 269

               Z = ( 269 – 269) / 20

Z = 0

              

For x = 271

               Z = ( 271 – 269) / 20

Z = 0.1

Probability between 269 and 271 = Area between Z=0 to Z=0.1

= 0.0398          ---------------- ( Referring to Z score table )

              Probability between 269 and 271 = 0.0398