Question

consider the distribution of monthly social security (OASDI) payments. Assume a normal distribution with a standard deviation of $116. if one-fourth of payments are above $1214,87 what is the mean monthly payment?

Answer 1

$1137

Explanation:

Solution:-

We will define the random variable as follows:

X: Monthly social security (OASDI) payments

The random variable ( X ) is assumed to be normally distributed. This implies that most monthly payments are clustered around the mean value ( μ ) and the spread of payments value is defined by standard deviation ( σ ).

The normal distribution is defined by two parameters mean ( μ ) and standard deviation ( σ ) as follows:

X ~ Norm ( μ , σ^2 )

We will define the normal distribution for (OASDI) payments as follows:

X ~ Norm ( μ , 116^2 )

We are to determine the mean value of the distribution by considering the area under neat the normal distribution curve as the probability of occurrence. We are given that 1/4 th of payments lie above the value of $1214.87. We can express this as:

P ( X > 1214.87 ) = 0.25

We need to standardize the limiting value of x = $1214.87 by determining the Z-score corresponding to ( greater than ) probability of 0.25.

Using standard normal tables, determine the Z-score value corresponding to:

P ( Z > z-score ) = 0.25 OR P ( Z < z-score ) = 0.75

z-score = 0.675

- Now use the standardizing formula as follows:

`z-score = (x - u)/(sigma) \\\\1214.87 - u = 0.675*116\\\\u = 1214.87 - 78.3\\\\u = 1136.57`

Answer: The mean value is $1137