the daily milk production for a certain type of dairy cow has a mean of 43 pounds per day with a standard diviation of 5 pounds per day. Suppose this daily mild production has a normal distribution. Calculate the probability that a randomly selected cow of this type has a daily mild production that is greater than 45 pounds.

either use a table, a statistical calculator, or here:http://davidmlane.com/hyperstat/z_table.html

27 percent?

To calculate the probability, we will use the concept of the standard normal distribution. However, in this case, we need to transform the values to the standard normal distribution using z-scores.

Step 1: Calculate the z-score
The z-score represents the number of standard deviations that a particular value is away from the mean. We can calculate the z-score using the formula:

z = (x - μ) / σ

Where:
x = the value we are interested in (45 pounds),
μ = the mean (43 pounds), and
σ = the standard deviation (5 pounds).

Substituting the values into the formula, we get:

z = (45 - 43) / 5 = 2 / 5 = 0.4

Step 2: Find the probability
Once we have the z-score, we can use a standard normal distribution table or a calculator to find the probability. The probability can be interpreted as the area under the curve to the right of the z-score.

For example, if we use a standard normal distribution table, we look for the value closest to 0.4. In this case, the closest value we can find is 0.40, which corresponds to a probability of approximately 0.6554.

Therefore, the probability that a randomly selected cow of this type has a daily milk production greater than 45 pounds is approximately 0.6554 or 65.54%.

Note: The standard normal distribution table provides values for the area under the curve to the left of the z-score. Since we are interested in the area to the right of the z-score, we subtract the value from 1.