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

Option C: 15.87 is the correct answer

Step-by-step 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:

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

your answer is the frist one

hop i

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497