Question

Household incomes are normally distributed with a mean of $47,500 and a standard deviation of $12,500. What is the probability of having incomes of $60,000

or more?
a) 2.50%
O b) 5.0096
C) 15.87%
od) 7.8096

Answer 1

Option C: 15.87 is the correct answer

Explanation:

Given

Mean = $47500

SD = $12500

We have to find the probability of having $60000 or more

So,

x = $60000

The z-score will be:

`z-score = z = (x-mean)/(SD)\\z = (60000-47500)/(12500)\\= (12500)/(12500)\\= 1`

Looking in the z-score table, we get

P(Z<60000) = 0.8413

In order to find the probability P(Z>60000) => 1-P(Z<60000)

= 1-0.8413

=0.1587

Converting into percentage: 0.1587*100 = 15.87%

Hence,

Option C: 15.87 is the correct answer