Question

The bursting strength of bottles of a certain type is normally distributed with μ=250 psi and σ= 30 psi. If these bottles are shipped 12 to a carton, in what proportion of cartons will at least one of the bottles have a bursting strength exceeding 300 psi? Hint: define a "success" as a bottle having a bursting strength exceeding 300 psi. Consider how you might determine the probability of "success."

Answer 1

4.75% of the bottles are succesful.

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 = 250, \sigma = 30`.

If these bottles are shipped 12 to a carton, in what proportion of cartons will at least one of the bottles have a bursting strength exceeding 300 psi?

This proportion is the percentage of the bootles that exceed 300 psi. This is 1 subtracted by the pvalue of Z when
`X = 300`.

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

`Z = (300 - 250)/(30)`

`Z = 1.67`

`Z = 1.67` has a pvalue of 0.9525.

This means that the proportion of sucesses is 1-0.9525 = 0.0475 = 4.75% of the bottles.