Question

A manufacturer produces gears for use in an engine’s transmission that have a mean diameter of 10.00 mm and a standard deviation of 0.03 mm. The length of these diameters follows the normal distribution. What is the probability that a randomly selected gear has a diameter between 9.96 mm and 10.01 mm?

Answer 1

Pr=0.2894

Explanation:

given mean diameter =10 mm

standard deviation=0.03 mm

z equation is

z=x-μ/σ

The problem has two values of x

for x=9.96

z=-1.33

for x-10.01

z=0.33

from Probability table we have

Pr(-1.33<z<0.33)=pr(z<0.33)-pr(z>-1.33)

Pr=0.2894

Answer 2

Answer: 0.2789

Explanation:

Given: Mean :
`\mu=10.00\ mm`

Standard deviation :
`\sigma =0.03\ mm`

The formula to calculate z-score is given by :_

`z=(x-\mu)/(\sigma)`

For x= 9.96 mm, we have

`z=(10-9.96)/(0.03)\approx1.33`

For x= 10.01 mm, we have

`z=(10.01-10)/(0.03)\approx0.33`

The P-value =
`P(0.33<z<1.33)=P(z<1.33)-P(z<0.33)`

`= 0.9082408- 0.6293=0.2789408\approx0.2789`

Hence, the probability that a randomly selected gear has a diameter between 9.96 mm and 10.01 mm =0.2789