Question

A civil service exam yields scores with a mean of 81 and a standard deviation of 5.5. Using Chebyshev's Theorem what can we say about the percentage of scores that are above 92?

Select one:

A. At most 12.5% of the scores are above 92.

B. At most 25% of the scores are above 92.

C. At least 75% of the scores are above 92.

D. At least 25% of the scores are above 92.

E. None of the above

Answer 1

Solution:

Given: A civil service exam yields scores with a mean of 81 and a standard deviation of 5.5.

Thus mean =
u= 81
and standard deviation =
45.5

We have to find the percentage of scores that are above 92 using Chebyshev's Theorem.

According to Chebyshev’s inequality, at most of the data fall outside k standard deviation from mean.

1

Thus find k:

92 is above 81 so use:

u + ko = Upper Limit

81 + k x 5.5 = 92

k x 5.5 = 92 - 81

k x 5.5 = 11

$k =\frac{11}{5.5}$

Thus 92 is 2 standard deviation above mean.

Thus find:

1 % = . ..

$\left ( \frac{1}{k^{2}} \right ) =\left ( \frac{1}{2^{2}} \right )$

$\left ( \frac{1}{k^{2}} \right ) =\left ( \frac{1}{4} \right )$

(三) -0.25

(三) - 25%。

At most 25% data fall outside 2 standard deviation , that is : below 2 standard deviation and above 2 standard deviation, but we have to find only % of data above 2 standard deviation, thus half of 25% = 12.5%

Thus correct option is:

A. At most 12.5% of the scores are above 92.