Question

The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 30 liters, and standard deviation of 5.2 liters. A) What is the probability that daily production is less than 21.8 liters? Answer= (Round your answer to 4 decimal places.) B) What is the probability that daily production is more than 21.3 liters? Answer= (Round your answer to 4 decimal places.)

Answer 1

Here we have given that

The below mentioned parameters is assumed to be normally distributed,

= The mean daily production of a herd of cows= 30 liters

= population standard deviation= 5.2 liters

(A)

Now we want to find the probability that daily production is less than 21.8 liters i.e. P(X < 21.8)

For that 1^st we want to find the Z-score

Z-score =

=

= -1.58

i.e we get P( Z < -1.58 ) = 0.0571 ( using z standard normal table)

We get

the probability that daily production is less than 21.8 liters i.e. P(X < 21.8) is 0.0571

(B)

Now we want to find the probability that daily production is more than 21.3 liters i.e. P(X < 21.3)

For that 1^st we want to find the Z-score

Z-score =

=

= -1.67

i.e we get P( Z > -1.67 ) = 1- 0.0475 ( using z standard normal table)

=0.9525

We get

the probability that daily production is more than 21.3 liters i.e. P(X < 21.3) is 0.9525