Question

The annual earnings for a random sample of employees with CPA certification and 6 years of experience and working for large firms have a bell-shaped distribution with a mean of $134,000and a standard deviation of $12,000

a) Using the empirical rule, find the percentage of all such employees whose annual earnings are between
(i) $98,000 and $170,000 ; (ii) $110,000$110,000 and $158,000

b) Using the empirical rule, find the interval that contains approximately the annual earnings of 68% of all such employees.

Answer 1

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The normal distribution parameters are given as:

Mean, Mu = $134000

Stdev, Sigma = $12000

We will use the standardization formula, Z = (X-Mean)/(Stdev/sqrt(n))

i. 98000 is 134000-3*12000, i.e. 98000 is 3 deviations less than the mean.

Also, 170000 is 3 134000+3*12000 = 170000 i.e. 3 deviations more than the mean.

So, percentage of all such employess whose annual earnings are between $98000 and $170000 is 99.7%

(empirical rule for normal distribution states that 3 standard deviation away from mean covers 99.7% of area under curve)

Answer: 99.7%

ii. 110000 and 158000 are 2 standard deviation away from mean.
So, percentage of all such employess whose annual earnings are between $110000 and $158000 is 95%

( empirical rule for normal distribution states that 2 standard deviation away from mean covers 95% of area under curve)

Answer: 95%

b) By empirical rule 68% of area under curve is covered by 1 standard deviation around mean. So, the interval that contains approximately the annual earnings of 68% of all such employees is 134000+/- 1*12000 = 122000 to 146000

Answer: $122000 to $146000