Answer:
75.2 pounds
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 85, \sigma = 5[/tex]
Calculate the value in pounds (lbs) for the weekly demand below which the manager will have to reduce the price.
Lowest 2.5%, which is the value of X when Z has a pvalue of 0.025. So X when [tex]Z = -1.96[/tex]
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.96 = \frac{X - 85}{5}[/tex]
[tex]X - 85 = -1.96*5[/tex]
[tex]X = 75.2[/tex]
