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when lavaughn commutes to work, the amount of time it takes him to arrive is normally distributed with a mean of 57 minutes and a standard deviation of 4.5 minutes. what is the probability that on a given day, his commute will be longer than 49 minutes, to the nearest thousandth?

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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