Question

The time for an oil change is normally distributed with mean μ = 11.5 minutes and standard deviation LaTeX: \sigma σ= 1.2 minutes. For what percentage of cars will the oil change take between 9 and 13 minutes?

Answer 1

Answer:

87.572%

Step-by-step explanation:

We start by calculating the z-scores for each of the minutes

Z-score = (x-mean)/SD

Where x is the raw score, mean = 11.5 and standard deviation is 1.2

For 9 minutes

z-score = (9-11.5)/1.2 = -2.08

For 13 minutes

z-score = (13-11.5)/1.2 = 1.25

So the probability we want to calculate is;

P( -2.083 < x < 1.25)

we can use the standard normal distribution table to get this

The probability will be;

P(x < 1.25) - P (-2.083< x) = 0.87572

This represents a percentage of 87.572%