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The manager of a healthy food store has determined that the weekly demand for a popular type of granola is a normally distributed random variable with mean 85 pounds and standard deviation 5 pounds. If the demand for a given week falls within the lowest 2.5% of all possible values for the weekly demand, the price of the granola will be reduced for the following week. Calculate the value in pounds (lbs) for the weekly demand below which the manager will have to reduce the price.

Respuesta :

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]

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