Question

# Assume the average age of an MBA student is 31.7 years old with a standard deviation...

Assume the average age of an MBA student is 31.7 years old with a standard deviation of 2.6 years.

​a) Determine the coefficient of variation.

​b) Calculate the​ z-score for an MBA student who is 26 years old.

​c) Using the empirical​ rule, determine the range of ages that will include 95​% of the students around the mean.

​d) Using​ Chebyshev's Theorem, determine the range of ages that will include at least 92​% of the students around the mean.

​e) Using​ Chebyshev's Theorem, determine the range of ages that will include at least 85​% of the students around the mean.

(a) Coefficient of Variation = (Standard Deviation / Mean) * 100

= (2.6 / 31.7) * 100 = 8.20

(b) The Z score for any sample value X is calculated as, Z = (X - μ) / σ , where X is sample data, μ is mean and σ is standard deviation

= (26 - 31.7) / 2.6 = -2.19

(c) According to the Empirical Rule (also called the 68-95-99.7 Rule), we would expect about 68% of all observations to fall within one standard deviation of the mean; about 95% to fall within two standard deviations of the mean; and about all (99.7%) to fall within three standard deviations of the mean.

Therefore, 95% would fall between 2 standard deviations from the mean
= μ +- 2*σ
= Between, 31.7 - (2 * 2.6) and 31.7 - (2 * 2.6)
= Between 26.5 and 36.9

(d)

Chebyshev's theorem states that, precentage of data will lie within k standard deviations of the mean.

Therefore for atleast 92% of students to lie around mean, (1 - 1/k2) * 100 = 92

Solving this for k, it gives k = 3.53, therefore 92% will lie within 3.53 standard deviations of mean

= 31.7 - (3.53 * 2.6)   and 31.7 + (3.53 * 2.6)

= Between 22.5 and 40.8 (rounded of to 1 decimal point)

(e) This part is same as above part (d) except it is to be calculated for 85%, so we will use same mentod and formula.

Therefore for atleast 85% of students to lie around mean, (1 - 1/k2) * 100 = 85

Solving this for k, it gives k = 2.58, therefore 85% will lie within 2.58 standard deviations of mean

= 31.7 - (2.58 * 2.6)   and 31.7 + (2.58 * 2.6)

= Between 25.0 and 38.4 (rounded of to 1 decimal point)

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