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16) Assuming that the number of shoppers who buy a packet of crackers is approximately normal with a mean of 20 and standard deviation of 4, what is the approximate probability that at least 28 of the next 100 shoppers who sample the crackers will buy a pack?

Respuesta :

Answer:

2.28% probability that at least 28 of the next 100 shoppers who sample the crackers will buy a pack

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 = 20, \sigma = 4[/tex]

What is the approximate probability that at least 28 of the next 100 shoppers who sample the crackers will buy a pack?

This probability is 1 subtracted by the pvalue of Z when X = 28. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{28 - 20}{4}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that at least 28 of the next 100 shoppers who sample the crackers will buy a pack