Question

The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 39 liters, and standard deviation of 11.6 liters.

A) What is the probability that daily production is between 42 and 68.7 liters? Do not round until you get your your final answer.

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.

Answer 1

To find the probability that the daily production of the cows is between 42 and 68.7 liters, we will use the properties of the normal distribution.

Given:
- Mean (\(\mu\)) = 39 liters
- Standard Deviation (\(\sigma\)) = 11.6 liters

### Step 1: Calculate the Z-scores for the values 42 and 68.7.

The Z-score formula is:

\[
Z = \frac{(X - \mu)}{\sigma}
\]

#### For \(X = 42\):

\[
Z_{42} = \frac{(42 - 39)}{11.6} = \frac{3}{11.6} \approx 0.2586
\]

#### For \(X = 68.7\):

\[
Z_{68.7} = \frac{(68.7 - 39)}{11.6} = \frac{29.7}{11.6} \approx 2.5593
\]

### Step 2: Use the Z-scores to find the probabilities.

You can use a calculator or statistical software to find the probabilities corresponding to these Z-scores.

- Let \(P(Z \leq Z_{42})\) be the cumulative probability for \(Z_{42}\):
\[
P(Z \leq 0.2586) \approx 0.6016
\]

- Let \(P(Z \leq Z_{68.7})\) be the cumulative probability for \(Z_{68.7}\):
\[
P(Z \leq 2.5593) \approx 0.9947
\]

### Step 3: Calculate the probability that production is between 42 and 68.7 liters.

\[
P(42 < X < 68.7) = P(Z \leq 2.5593) - P(Z \leq 0.2586)
\]

\[
P(42 < X < 68.7) \approx 0.9947 - 0.6016 = 0.3931
\]

### Step 4: Round the answer to four decimal places.

The probability that daily production is between 42 and 68.7 liters is:

\[
\text{Answer} = 0.3931
\]