Question

The mean household income in a country in a recent year was about ​$77 comma 044 and the standard deviation was about ​$84 comma 000. ​(The median income was ​$58 comma 423​.) ​a) If a Normal model was used for these​ incomes, what would be the household income of the top 1​%? ​b) How confident should one be in the answer in part​ a? ​c) Why might the Normal model not be a good one for​ incomes?

Answer 1

Answer:

a) Income of $272,428 or more would be top 1%.

b) Skewed right

c) Not always normally distributed

Explanation:

We are given the following information in the question:

Mean, μ = $77,044

Standard Deviation, σ = $84,000

Median = $58,423

a) We follow a normal mode

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.01

P(X > x)  

$P( X > x) = P( z > \displaystyle\frac{x - 77044}{84000})=0.03$  

$= 1 -P( z \leq \displaystyle\frac{x - 77044}{84000})=0.01$  

$=P( z \leq \displaystyle\frac{x - 77044}{84000})=0.99$  

Calculation the value from standard normal z table, we have,  

$\displaystyle\frac{x - 77044}{84000} = 2.326\\\\x = 272428$

Thus, income of $272,428 or more would be top 1%.

b) We should not be confident as the median is not equal to the mean. Hence, it is not a normal distribution. It was just an assumption. Since the mean is greater than the median the distribution of income is skewed towards right.

c) Normal model not be a good one for​ incomes because the median may not always e equal to the mean and hence, they do not follow a normal distribution.