Suppose that 25% of horses live over 23.4 years, while 85% live less than 25.2 y

Question

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

P(X > 23.4) = 0.25

Z(X > 23.4) = 0.6745

P(X < 25.2) = 0.85

Z(X < 25.2) = 1.0365

=> + 0.6745 = 23.4

+ 1.0365 = 25.2

Subtracting

=> - 0.3620 = -1.8

=> = 1.8 / 0.3620 = 4.9724

Substituting in the first equation

=> + 0.6745 * 4.9724 = 23.4

=> = 23.4 - 0.6745 * 4.9724

= 20.05.

The nearest answer is C.mean 20.045; standard deviation 4.974.

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