Question

The diameters of a batch of ball bearings are known to follow a normal distribution with a mean 4.0 in and a standard deviation of 0.15 in. If a ball bearing is chosen randomly, find the probability of realizing the following event: (a) a diameter between 3.8 in and 4.3 in, (b) a diameter smaller than 3 9 in, (c) a diameter larger than 4.2 in

Answer 1

a) 88.54% probability of a diameter between 3.8 in and 4.3 in

b) 25.14% probability of a diameter smaller than 3.9in

c) 90.82% probability of a diameter larger than 4.2 in

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 4, \sigma = 0.15`

(a) a diameter between 3.8 in and 4.3 in,

This is the pvalue of Z when X = 4.3 subtracted by the pvalue of Z when X = 3.8.

X = 4.3

`Z = (X - \mu)/(\sigma)`

`Z = (4.3 - 4)/(0.15)`

`Z = 2`

`Z = 2` has a pvalue of 0.9772

X = 3.8

`Z = (X - \mu)/(\sigma)`

`Z = (3.8 - 4)/(0.15)`

`Z = -1.33`

`Z = -1.33` has a pvalue of 0.0918

0.9772 - 0.0918 = 0.8854

88.54% probability of a diameter between 3.8 in and 4.3 in

(b) a diameter smaller than 3 9 in,

This is the pvalue of Z when X = 3.9. So

`Z = (X - \mu)/(\sigma)`

`Z = (3.9 - 4)/(0.15)`

`Z = -0.67`

`Z = -0.67` has a pvalue of 0.2514

25.14% probability of a diameter smaller than 3.9in

(c) a diameter larger than 4.2 in

This is 1 subtracted by the pvalue of Z when X = 4.2. So

`Z = (X - \mu)/(\sigma)`

`Z = (4.2 - 4)/(0.15)`

`Z = 1.33`

`Z = 1.33` has a pvalue of 0.9082

90.82% probability of a diameter larger than 4.2 in