Question

The weekly demand for a product has a normal distribution with a mean of 1,000 and a standard deviation of 200. The current on-hand inventory is 750 and no deliveries will be occurring in the next week

Answer 1

Using the normal distribution, it is found that there is a 0.8943 = 89.43% probability of a demand above 750.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

In this problem:

• The mean is of 1000, hence $\mu = 1000$.

• The standard deviation is of 200, hence $\sigma = 200$

The probability of a demand above 750 is one subtracted by the p-value of Z when X = 750, hence:

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

$Z = \frac{750 - 1000}{200}$

$Z = -1.25$

$Z = -1.25$ has a p-value of 0.1057.

1 - 0.1057 = 0.8943

0.8943 = 89.43% probability of a demand above 750.

A similar problem is given at https://brainly.com/question/24663213