The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 31 liters, and standard deviation of 7.2 liters.

Question

A.) What is the probability that daily production is less than 29.7 liters ?

B.) What is the probability that daily production is more than 51.8 liters ?

Answer 1

The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 31 liters, and standard deviation of 7.2 liters.

A.) What is the probability that daily production is less than 29.7 liters?

B.) What is the probability that daily production is more than 51.8 liters?

Step-1

Given

Population is normally distributed

Population mean m=31

Population Standard deviation s=7.2

Step-2

Compute z score of 29.7: (29.7-m)/s

z(29.7) = (29.7-31)/7.2 = -0.181

Look up the z table and find area under the curve left of z =  -0.181

Probability = 0.42836

(A) Answer: probability that daily production is less than 29.7 liters = 0.42836

Step-3

Compute z score of 51.8: (51.8-m)/s

z(51.8) = = (51.8-31)/7.2 = 2.89

Look up the z table and find area under the curve right of z = 2.89

Probability =  0.001933

(B) Answer: probability that daily production is more than 51.8 liters = 0.001933