Respuesta :
Answer:
94.26% probability that a randomly selected individual will have a waiting time between 15 and 45 minutes.
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 = 30, \sigma = 8[/tex]
What is the probability that a randomly selected individual will have a waiting time between 15 and 45 minutes?
This is the pvalue of Z when [tex]X = 45[/tex] subtracted by the pvalue of Z when [tex]X = 15[/tex].
So
X = 45
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{45 - 30}{8}[/tex]
[tex]Z = 1.9[/tex]
[tex]Z = 1.9[/tex] has a pvalue of 0.9713.
X = 15
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{15 - 30}{8}[/tex]
[tex]Z = -1.9[/tex]
[tex]Z = -1.9[/tex] has a pvalue of 0.0287.
So there is a 0.9713 - 0.0287 = 0.9426 = 94.26% probability that a randomly selected individual will have a waiting time between 15 and 45 minutes.
