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

Answer:

4.75% of the bottles are succesful.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 $X = 300$.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{300 - 250}{30}$

$Z = 1.67$

$Z = 1.67$ has a pvalue of 0.9525.

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