The per capita disposable income for residents of a U.S. city in a recent year is normally distributed, with a mean of about $44,000 and a standard deviation of about $2450. Between what two values does the middle 60% of disposable income lie? Round to the nearest hundredth and list values from least to greatest.

 

The middle 60% of values is between 41,942 and 46,058.

Step-by-step explanation

If you are looking for the middle 60% of data, this means that you're looking for the data that is 30% above the mean and 30% below the mean. Because the mean is represented by p= 0.5, add and subtract 0.3 from 0.5.

0.5 + 0.3 = 0.8

0.5 - 0.3 = 0.2

These are the p values that you want to find corresponding values for.

Now, you need to look in the standard normal table to find the corresponding z scores for when p = 0.8 and p = 0.2.

When p = 0.2, z = -0.84.

When p = 0.8, z = 0.84.

Plug both of these values into the following equation, with the mean and standard deviation provided in the problem, to solve for x.

z = (x - mean)/ standard deviation

-0.84 = (x - 44,000)/2450

-2058 = x - 44,000

x = 41,942

0.84 = (x - 44,000)/2450

2058 = (x - 44000)

x = 46,058

The middle 60% of values is between 41,942 and 46,058.