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Question

Nominal income refers to an income value that is not adjusted for inflation. Real income adjusts the nominal value for the rate of inflation. Real income per capita is considered a good measure of the buying power of an individual in a particular area because it adjusts for both the number of people in the area and the relative inflation of the area. According to Forbes magazine, the state with the highest annual real income per capita in 2020 is Connecticut with a real per capita income of $68,533. Suppose that the annual real income of individuals in Connecticut follows a normal distribution with a mean of $68,533 and standard deviation of 23.480

a. What is the probability that an individual in Connecticut has an annual real income of $50,000 or more (to 2 decimals)?

c. What is the probability that an individual in Connecticut has an annual real income between$60,000 and $80,000 (to 2 decimals)?

 

d. What is the annual real income of a person in the 99th percentile of annual real income in Connecticut (to 2 decimals)?

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Answer 1

Hi Christina,

a. This is the classic equation in introductory statistics: z= (x-mu)/sigma

x= Income you are interested in

mu= Population Mean Income

sigma= standard deviation

Here,

x= 50000

mu= 68533

sigma= 23480

z= (50000-68533)/23480

z= -0.79

From z-table, we can obtain the probability that z<-0.79

P(Z< -0.79) = P(X<50000) = 0.21

But the question calls for greater than 50000, so we apply the Complement Rule:

P(Z>a) = 1 - P(Z<a); a= any real number

P(Z> -0.79) = P(X>50000) = 0.79

c. Same formula, but this time, we need to do it twice since we have two x-values, so i'll use z, x1, z2, and x2:

x1= 60000

mu= 68533

sigma= 23480

z1 = (60000-68533)/23480

z1= -0.36

Keep that z-score in mind; we need to find z2:

x2= 80000

mu= 68533

sigma= 23480

z2= (80000-68533)/23480

z2= 0.49

Now, to get the probability that x falls between those two salaries, we go to the z-table, get the probability for z<z1 and z<z2. We then subtract those probabilities. Thus:

P(Z<-0.36) = 0.36

P(Z<0.49) = 0.69

P(-0.36<z<0.49) = P(60000<x<80000) = 0.33

d. For this, it's the same formula, but we need to look at the interior of the z-table first to get z at the 99th percentile, which means you are looking in the interior for the closest value to 0.99:

z_.99= 2.33 (approximately)

So:

z= 2.33

mu= 68533

sigma= 23480

Solve for x:

2.33= (x-68533)/23480

Multiply both sides by 23480:

54708.4= x - 68533

Add 68533 to both sides:

x= $123241.40