Assume that the speed of automobiles on an expressway during rush hour is normally distributed with a mean of 70 mph and a standard deviation of 7

mph. What percent of cars are traveling faster than 70 ​mph?

You can play around with Z table stuff here:

http://davidmlane.com/hyperstat/z_table.html

To find the percentage of cars traveling faster than 70 mph, we need to calculate the area under the normal distribution curve to the right of 70 mph. This represents the proportion of cars with a speed greater than 70 mph.

First, we need to standardize the value of 70 mph by converting it to a z-score using the formula:

z = (x - μ) / σ

Where:
- x is the value we want to standardize (70 mph in this case),
- μ is the mean of the distribution (70 mph),
- σ is the standard deviation of the distribution (7 mph).

Plugging in the values:

z = (70 - 70) / 7
z = 0 / 7
z = 0

The z-score is 0. This means that the value of 70 mph is exactly at the mean of the distribution.

Next, we need to find the proportion of the distribution that lies to the right of the z-score of 0. We can use a z-table or a statistical calculator to find this value.

Looking up the z-score of 0 in a z-table, we find that the proportion of the distribution to the right of z = 0 is about 0.5 or 50%.

Therefore, approximately 50% of cars are traveling faster than 70 mph on the expressway during rush hour.