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Sagot :
Using the normal distribution, it is found that there is a 0.963 = 96.3% probability that on a given day, his commute will be longer than 49 minutes.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, the mean and the standard deviation are given, respectively, by:
[tex]\mu = 57, \sigma = 4.5[/tex].
The probability that on a given day, his commute will be longer than 49 minutes is one subtracted by the p-value of Z when X = 49, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{49 - 57}{4.5}[/tex]
Z = -1.78
Z = -1.78 has a p-value of 0.0375.
1 - 0.0375 = 0.963.
0.963 = 96.3% probability that on a given day, his commute will be longer than 49 minutes.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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