Solving Word Problems Involving Chebyshev's Theorem

Solving Word Problems Involving Chebyshev’s Theorem

Chebyshev’s Theorem states that the proportion or percentage of any data set that lies within k standard deviation of the mean where k is any positive integer greater than I is at least 1 – 1/k^2.

Sample Problem Number One :

The mean score of Insurance Commission Licensure Examination is 75 with a standard deviation of 5. What percentage of the data set lies between 50 and 100 ?

Solution : First find the value of k

Mean – (k) (sd) = lower limit

75 – 5K - 50

75 – 50 = 5k

25 = 5k

K = 5

To get the percentage use 1 – 1/k^2

1 - 1/ 25 = 24/25 = 96%

96% of the data set lies between 50 and 100.

Sample Problem Number Two :

The mean age of flight attendant of PAL is 40 years old with a standard deviation of 8. What percent of the data set lies between 20 and 60 ?

Solution : First find the value of k

40 – 20 = 8k

20 = 8k

k = 2.5

To find the percentage : 1 – 1/(2.5)^2 = 84 %

84% of the data set lies between the ages 20 and 60.

Sample Problem Number Three :

The mean age of saleslady in ABC Dept Store is 30 with a standard deviation of 6 . Between which two age limit must 75% of the data set lie ?

Solution : First find the value of k

1 - 1/k^2 = ¾

1 - ¾ = 1/k^2

¼ = 1/k^2

k^2 = 4

k = 2

Lower age limit :

30 - (k ) (sd) = 30 - (6)(2) = 30 -12 = 18

Upper age limit :

30 + ( k) (sd) = 30 + (6)(2) = 30 + 12 = 42

The mean age of 30 with an sd of 6 must lie between 18 and 42 to represent 75% of the data set.

Sample Problem Number Four :

The mean score in an accounting test is 80 with a standard deviation of 10. Between which two scores must this mean lie to represent 8/9 of data set ?

Solution : Find first the value of k

1 - 1 /k^2 = 8/9

1 - 8/9 = 1/k^2

1/9 = 1/k^2

k^2 = 9

k = 3

Lower limit :

80 – (10)(3) = 80 – 30 = 50

Upper limit :

80 + 30 = 110

The mean score of 60 with an sd of 10 must lie between 50 and 110 to represent 88.89% of the data set.