A company is offering 401k matching for its employees who stay with the company for more than 10 years. The company's CFO finds that the average retirement account holds $490,000, with a standard deviation of $55,000, distributed normally. What amount of money separates the lowest 20% of the means of retirement accounts from the highest 80% in a sampling of 80 employees

Given Information:

Mean of retirement accounts = μ = $490,000

Standard deviation of retirement accounts = σ =  $55,000

Sample size = n = 80

$484,835 is the amount that separates the lowest 20% of the means of retirement accounts from the highest 80% of the means of retirement accounts.

Step-by-step explanation:

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.  

The amount of money that separates the lowest 20% of the means of retirement accounts from the highest 80% is given by

The z-score corresponding to 0.20 is -0.84

Rounding off to the nearest whole number

Therefore, $484,835 is the amount that separates the lowest 20% of the means of retirement accounts from the highest 80% of the means of retirement accounts.

How to use z-table?

In the z-table find the probability of 0.20

Note down the value of that row, it would be -0.8.

Note down the value of that column, it would be 0.04.

So the z-score is -0.84

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!