Question

The weights of turkeys are normally distributed with a mean of 22.5 pounds and a standard deviation of 4.3 pounds. Two turkeys are chosen at random. Assuming independent events, what is the probability that the combined weights of the turkeys will be more than 40 pounds

Answer 1

Answer:

0.9495 = 94.95% probability that the combined weights of the turkeys will be more than 40 pounds.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If we add n occurences of a normal variable, we have that:

$\mu = \mu*n, \sigma = \sqrt{n}\sigma$

In this question, we have that:

$\mu = 22.5, \sigma = 4.3$

Two turkeys, so:

$\mu = 22.5*2 = 45, \sigma = 4.3\sqrt{2} = 6.08$

What is the probability that the combined weights of the turkeys will be more than 40 pounds?

This is one subtracted by the pvalue of Z when X = 40. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{40 - 45}{6.08}$

$Z = -1.64$

$Z = -1.64$ has a pvalue of 0.0505

1 - 0.0505 = 0.9495

0.9495 = 94.95% probability that the combined weights of the turkeys will be more than 40 pounds.