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 5.7 liters.

A) What is the probability that daily production is between 21.5 and 40.3 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.

The answer is not 0.8960

1 answer

To find the probability that the daily production of a herd of cows is between 21.5 and 40.3 liters, we'll use the properties of the normal distribution. Specifically, we will standardize the values to find their corresponding Z-scores and then utilize the cumulative distribution function (CDF) for the standard normal distribution.

Step 1: Calculate the Z-scores.

The formula to convert a raw score \(X\) to a Z-score is:

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

Where:

  • \(X\) is the raw score,
  • \(\mu\) is the mean (30 liters),
  • \(\sigma\) is the standard deviation (5.7 liters).

For \(X = 21.5\) liters:

\[ Z_{21.5} = \frac{21.5 - 30}{5.7} = \frac{-8.5}{5.7} \approx -1.4912 \]

For \(X = 40.3\) liters:

\[ Z_{40.3} = \frac{40.3 - 30}{5.7} = \frac{10.3}{5.7} \approx 1.8035 \]

Step 2: Find probabilities using the CDF.

To find the probability of being between these two Z-scores, we will look for the cumulative probabilities for \(Z_{21.5}\) and \(Z_{40.3}\).

We can use a computational tool (like Python, R, or a scientific calculator) to find the CDF values directly for the Z-scores.

Using the cumulative distribution function \( \Phi(z) \):

  • For \(Z = -1.4912\): \( \Phi(-1.4912) \)
  • For \(Z = 1.8035\): \( \Phi(1.8035) \)

Using a standard normal distribution calculator or a statistical software:

\[ \Phi(-1.4912) \approx 0.0685 \] \[ \Phi(1.8035) \approx 0.9640 \]

Step 3: Calculate the probability between the two Z-scores.

The probability \(P(21.5 < X < 40.3)\) is given by:

\[ P(21.5 < X < 40.3) = \Phi(1.8035) - \Phi(-1.4912) \]

Substituting the values:

\[ P(21.5 < X < 40.3) \approx 0.9640 - 0.0685 = 0.8955 \]

Final Answer (rounded to 4 decimal places):

\[ \text{Probability} = 0.8955 \]

So the final answer is:

Answer = 0.8955