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Sagot :
Using the normal distribution, there is a 0.1587 = 15.87% probability that’s a car picked at random is traveling at more than 100 km/hr.
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.
The mean and the standard deviation are given, respectively, by:
[tex]\mu = 90, \sigma = 10[/tex]
The probability that’s a car picked at random is traveling at more than 100 km/hr is one subtracted by the p-value of Z when X = 100, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{100 - 90}{10}[/tex]
Z = 1
Z = 1 has a p-value of 0.8413.
1 - 0.8413 = 0.1587.
0.1587 = 15.87% probability that’s a car picked at random is traveling at more than 100 km/hr.
More can be learned about the normal distribution at https://brainly.com/question/28096232
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