The average commute to work (one way is 25 minutes according to the 2005 American Community Survey. If we assume that commuting times are normally distributed and that the standard deviation is 6.1 minutes, calculate the probability that a randomly selected commuter spends for the following cases. Round the final answers to four decimal places and intermediate z-value calculations to two decimal places. More than 31 minutes commuting one way P(X > 31) = Less than 8 minutes commuting one way P(X < 8)

The z-score of a measure X of a variable that has mean symbolized by [tex]\mu[/tex] and standard deviation symbolized by [tex]\sigma[/tex] is obtained by the rule presented as follows:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The z-score represents how many standard deviations the measure X is above or below the mean of the distribution, depending if the obtained z-score is positive or negative.

Using the z-score table, the p-value associated with the calculated z-score is found, and it represents the percentile of the measure X in the distribution.

The mean and the standard deviation of the commute times are given as follows:

[tex]\mu = 25, \sigma = 6.1[/tex]

The probability that the time is more than 31 minutes is one subtracted by the p-value of Z when X = 31, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Z = (31 - 25)/6.1

Z = 0.98

Z = 0.98 has a p-value of 0.8365.

1 - 0.8365 = 0.1635 = 16.35%.

The probability that the time is less than 8 minutes is the p-value of Z when X = 8, hence:

Z = (8 - 25)/6.1

Z = -2.79

Z = -2.79 has a p-value of 0.0026.

More can be learned about the normal distribution at https://brainly.com/question/25800303

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