Question

Assume that the speed of automobiles on an expressway during rush hour is normally distributed with a mean of 63 mph and a standard deviation of 10 mph. what percent of cars are traveling faster than 79 ​mph?

Answer 1

The speed is normally distributed with mean μ=63 and standard deviation σ =10.

Use the transformation $\frac{X-\mu}{\sigma}$ for calculating the probability. This value is called Z-score.

The Z score is $Z=\frac{79-63}{10} =1.6$.

Refer to the standard normal distribution table.

The probability

$P(X>79)=P(Z>1.6)=1-P(Z<1.6)=0.0548$

Thus, $5.48\%$ of the cars are travelling faster than 79 mph.