Respuesta :
Using the normal distribution, there is a 0.209 = 20.9% probability that the commute on a particular game day exceeds the commute on a particular non-game day.
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.
Researching the problem on the internet, on game days, the parameters are:
[tex]\mu = 20, \sigma = 9[/tex]
On non-game days, the parameters are:
[tex]\mu = 12, \sigma = 4[/tex]
For the distribution of differences, the mean and the standard deviation are given by:
- [tex]\mu = 12 - 20 = -8[/tex].
- [tex]\sigma = \sqrt{4^2 + 9^2} = 9.85[/tex]
The desired probability is P(X > 0), which is one subtracted by the p-value of 0, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (0 - (-8))/9.85
Z = 0.81
Z = 0.81 has a p-value of 0.7910
1 - 0.7910 = 0.209 = 20.9% probability that the commute on a particular game day exceeds the commute on a particular non-game day.
More can be learned about the normal distribution at https://brainly.com/question/24537145
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