Question

A pretzel company calculated that there is a mean of 73.5 broken pretzels in each production run with a standard deviation of 5.1. If the distribution is approximately normal, find the probability that there will be fewer than 69 broken pretzels in a run.

Answer 1

Answer:

0.1894 = 18.94% probability that there will be fewer than 69 broken pretzels in a run.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

A pretzel company calculated that there is a mean of 73.5 broken pretzels in each production run with a standard deviation of 5.1.

This means that $\mu = 73.5, \sigma = 5.1$

Find the probability that there will be fewer than 69 broken pretzels in a run.

This is the p-value of Z when X = 69.

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

$Z = \frac{69 - 73.5}{5.1}$

$Z = -0.88$

$Z = -0.88$ has a p-value of 0.1894

0.1894 = 18.94% probability that there will be fewer than 69 broken pretzels in a run.