Assume the average age of an MBA student is 29.5 years old with a standard devia

Question

Assume the average age of an MBA student is 29.5 years old with a standard deviation of 2.4 years.

a) The coefficient of variation is ____%
(Round to one decimal place as needed.)

b) The z-score for an MBA student who is 30 years old is ____.
(Round to two decimal places as needed.)

c) Using the emperical rule, the range of ages that will include 68% of the students around the mean, in interval notation, is (___ , ___).
(Round to one decimal place as needed.)

d) Using the Chesbyshev's Theorem, the range of ages that will include at least 92% of the students around the mean, in interval notation, is (___ , ___).
(Round to one decimal place as needed.)

e) Using the Chesbyshev's Theorem, the range of ages that will include at least 84% of the students around the mean, in interval notation, is (___ , ___).
(Round to one decimal place as needed.)

Explanation / Answer

a.
Mean ( u ) =29.5
Standard Deviation ( sd )=2.4
Normal Distribution = Z= X- u / sd ~ N(0,1)                  

coeffcient of variation = u/sd * 100% = (29.5/2.4) = 1229.17 ~ 12.2917%
b.
z-score when x=30,
P(X = 30) = (30-29.5)/2.4
= 0.5/2.4= 0.2083
c.
Given that the mean (u) of a normal probability distribution is 29.5 and the
standard deviation (sd) is 2.4                  
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e. (u ± 1s.d)
So to the given normal distribution about 68% of the observations lie in between
= [29.5 ± 2.4]
= [ 29.5 - 2.4 , 29.5 + 2.4]
= [ 27.1 , 31.9 ]  

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