Respuesta :
Answer:
Her commute would be between 32 and 35 minutes 33 times.
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 = 28, \sigma = 4.5[/tex]
Proportion of days in which the commute is between 32 and 35 minutes:
This is the pvalue of Z when X = 35 subtracted by the pvalue of Z when X = 32.
X = 35
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{35 - 28}{4.5}[/tex]
[tex]Z = 1.555[/tex]
[tex]Z = 1.555[/tex] has a pvalue of 0.94.
X = 32
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{32 - 28}{4.5}[/tex]
[tex]Z = 0.89[/tex]
[tex]Z = 0.89[/tex] has a pvalue of 0.8133.
0.94 - 0.8133 = 0.1267
Out of 262 days:
Each day, 0.1267 probability
0.1267*262 = 33
Her commute would be between 32 and 35 minutes 33 times.
