Question

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 3 liters.
a) what is the probability that daily production is less than 30.9 liters?
answer=
(round your answer to 4 decimal places.)
b) what is the probability that daily production is more than 35.7 liters?
answer=
(round your answer to 4 decimal places.)
warning: do not use the z normal tables...they may not be accurate enough since wamap may look for more accuracy than comes from the table.
check answer
question 17
4 pts details
score on last try: 4 of 4 pts. see details for more.

Explanation:

Step1: Calculate the z - score for part (a)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 30.9\), \(\mu=32\), and \(\sigma = 3\).

$$z=\frac{30.9 - 32}{3}=\frac{-1.1}{3}\approx - 0.37$$

Then, we find \(P(X\lt30.9)=P(Z\lt - 0.37)\). Using a calculator or software (not tables as per the warning), \(P(Z\lt - 0.37)\approx0.3557\)

Step2: Calculate the z - score for part (b)

Using the z - score formula \(z=\frac{x-\mu}{\sigma}\), with \(x = 35.7\), \(\mu = 32\), and \(\sigma=3\)

$$z=\frac{35.7-32}{3}=\frac{3.7}{3}\approx1.23$$

Then, \(P(X\gt35.7)=1 - P(X\leq35.7)=1 - P(Z\leq1.23)\). Using a calculator or software, \(P(Z\leq1.23)\approx0.8907\), so \(P(X\gt35.7)=1 - 0.8907 = 0.1093\)

Answer:

a) \(0.3557\)
b) \(0.1093\)