Question

The manager of spiffy lube auto lubrication shop is trying to revise his policy on ordering grease gun cartridges. currently he orders 110 cartridges per week, but he runs out of cartridges 1 out of 4 weeks. he knows that, on average, the shop uses 95 cartridges per week. he was also willing to assume that the demand for cartridges is normally distributed. a. compute the standard deviation of the distribution?

Answer 1

Using the normal distribution, it is found that the standard deviation is of 26.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

For this problem, we have that the mean is of $\mu = 95$, while X = 110 has a p-value of 0.75, hence X = 110, Z = 0.575, and then we solve for $\sigma$ to find the standard deviation.

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

$0.575 = \frac{110 - 95}{\sigma}$

$0.575\sigma = 15$

$\sigma = \frac{15}{0.575}$

$\sigma = 26$

More can be learned about the normal distribution at https://brainly.com/question/4079902

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