(Multiple Choice) Suppose that 25% of horses live over 23.4 years, while 85% liv

Question

(Multiple Choice) Suppose that 25% of horses live over 23.4 years, while 85% live less than 25.2 years. Assuming the ages of horses are normally distributed, what are the mean and standard deviation for the life expectancy of horses? Explain your answer! (Hint: Use 4 decimal places throughout)

A. mean 24.110; standard deviation 1.052

B. mean 22.690; standard deviation 1.052

C. mean 20.045; standard deviation 4.974

D. mean 22.690; standard deviation 4.974

E. Not enough information is given to find the mean and standard deviation.

Explanation / Answer

If 25% of horses live over 23.4 years, then 75% of horses live under 23.4 years. From the standard normal table, the z-score associated with 0.75 is 0.67. The z-score associated with 85% living less than 25.2 years is 1.04. Using the formula for standardized scores,
0.6745 = (23.4-mu)/sigma and 1.036 = (25.2-mu)/sigma
0.6745 sigma = 23.4-mu and 1.036 sigma = 25.2-mu
Now
1.036 sigma = 25.2-mu
0.6745 sigma = 23.4-mu
(-) (-) (+)
---------------------
0.3615sigma = 1.8
sigma = 4.79725
1.04 * 4.79725 = 25.2- mu  
mu = 20.0215

Correct answer: option (C) mean 20.045; standard deviation 4.974

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.