Question

Assume the average age of an MBA student is 29.1 years old with a standard deviation of 2.7 years a) Determine the coefficient of variation. b) Calculate the 2-score for an MBA student who is 24 years old c)Using the empirical rule, determine the range of ages that will include 997% of the students around the mean. d) Using Chebyshevs Theorem determine the range of ages that ill indude at least 96% of the students around the mean. e) Using Che yshe's Theorem, determne the range of ages that wil include at least 80% of the students around the mean a) The coeficent of variation is Round to one decimal place as needed) b) The z-score for an MBA student who is 24 years old is (Round to two decimal places as needed.) c Using the em cal rule, the rarge of ages that wil Round to one decimal place as needed ) de 97% of the stu nts around the men in interval n at is d Us n Chebys esT ore the range of ages that will ride at least Round to one decimal place as needed.) %ofthe stud ts ar the men in te val n tat is e Using Cheb stens The ore the rar eofa es that will na de at least 80% of the students ar the meanininterval notati is Enter your answer in each of the answer boxes O Type here to search Lenovo 2 3 4 6 Q W E R T Y U

Answer 1

Mean =29.1 , Standard Deviation = 2.7

A. Coefficient of variation = Standard Deviation / Mean = 2.7/29.1 = 0.093 = 9.28%

B. Age of MBA student =24 yrs

Z Score = (x- mean) / Standard Deviation = (24-29.1)/2.7 = -1.888

C. 99.7% on both sides conform to a Z score of 2.96

Therefore, the range will be ( Mean - Z Score * Std Dev , Mean + Z Score * Std Dev) = ( 29.1 - 2.96* 2.7 , 29.1 + 2.96* 2.7)

= ( 21.108, 37.092)

D. According to Chebyshev's Theorem, atleast (1-1/K²)*100 %of data lies within (Mean - K* Std.Dev) & (Mean + K* Std.Dev)

In this case, (1 - 1/K²)*100 = 96... => K = 5

The lower limit is Mean - K * Std. Dev = 29.1 - 5*2.7 = 15.6

The Upper limit is Mean + K * Std. Dev = 29.1 + 5*2.7 = 42.6

Therefore according to Chebyshev's Theorem , 96% of the values lies within (15.6,42.6)