Respuesta :
Answer:
4.55% probability that a randomly selected individual will take less than 5 minutes to select a shoe purchase.
Since Z > -2 and Z < 2, this outcome is not considered unusual.
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.
If [tex]Z \leq 2[/tex] or [tex]Z \geq 2[/tex], the outcome X is considered to be unusual.
In this question:
[tex]\mu = 8.21, \sigma = 1.9[/tex]
Find the probability that a randomly selected individual will take less than 5 minutes to select a shoe purchase.
This is the pvalue of Z when X = 5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5 - 8.21}{1.9}[/tex]
[tex]Z = -1.69[/tex]
[tex]Z = -1.69[/tex] has a pvalue of 0.0455.
4.55% probability that a randomly selected individual will take less than 5 minutes to select a shoe purchase.
Since Z > -2 and Z < 2, this outcome is not considered unusual.
