The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 32 liters, and standard deviation of 6.6 liters.

Question

A) What is the probability that daily production is less than 18.2 liters?

B) What is the probability that daily production is more than 17.2 liters?

I am unsure of what excel formula to use

Answer 1

x = daily production is N(32,6.6)

want to find P(x < 18.2)

transform to z = (x - mean)/SD which is N(0,1)

compute z for x = 18.2, mean = 32 and SD 6.6

z = (x - mean)/SD = (18.2-32)/6.6 = -2.09

P(x < 18.2) = P(z < -2.09) = 0.0183

want to find P(x > 17.2)

transform to z:

(17.2 - 32)/6.6 = -2.24

P(x > 17.2) = P(z > -2.24) = 1 - P(z < -2.24) = 1= 0.0125 = 0.9875