Question

A manufacturer contracts to supply ball bearings with diameters between 24.60 millimeters and 25.40 millimeters. Product analysis indicates that the ball bearings manufactured have diameters that are normally distributed with a mean of 25.10 millimeters and a standard deviation of 0.20 millimeters. What percentage of ball bearings fail to satisfy the contract specifications?

Answer 1

Answer: The percentage of ball bearings fail to satisfy the contract specifaction is 7.3 %.

Explanation:

First we need to find the standard score of 24.60 and 25.40. Using Standard Score formula.

`(24.60 - 25.10)/(.20) =-2.5\\\\(25.4 - 25.10)/(.20)=1.5`

now we need to find the percentages of these Z scores.

-2.5 = .0062 or 00.62%

1.5 = .9332 or 93.32%

now we subtract these two to get the in between percentage.

.9332 - .0062 = .927 or 92.7% is what doesnt fail to satifsy the contract spefications! But to find what does fail to satisfy we have to subtract it by 100% or 1.

1 - .927 = 0.073

7.3%

Answer 2

7.3%

Explanation:

to get this percentage, we need to use the z-score calculation;

z-score = (x-mean)/SD

mean = 25.10 , SD = 0.2

For diameter 24.60

z-score = (24.6-25.10)/0.2 = -0.5/0.2 = -2.5

For diameter 25.40

z-score = (25.4-25.1)/0.2 = 1.5

So the proportion that will satisfy the specifications will be;

P(-2.5 < z < 1.5)

At this point, we use the standard normal table

P(-2.5 < z < 1.5) = P (z<1.5) - P (Z < -2.5)

From standard normal table;

P(Z < 1.5) = 0.9332

P(z < -2.5) = 0.0062

P(-2.5 < z < 1.5) = 0.9332 - 0.0062 = 0.927

So the proportion that meets specification = 92.7% ( same as 0.927)

Proportion failing to meet specification = 100- 92.7% = 7.3%