Respuesta :
Answer:
$1137
Step-by-step 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:
[tex]z-score = \frac{x - u}{sigma} \\\\1214.87 - u = 0.675*116\\\\u = 1214.87 - 78.3\\\\u = 1136.57[/tex]
Answer: The mean value is $1137