Question

The weight of turkeys is normally distributed with a mean of 22 pounds and a standard deviation of 5 pounds. a. Find the probability that a randomly selected turkey weighs between 20 and 26 pounds. Your answer is . Round to 3 decimals and keep '0' before the decimal point. b. Find the probability that a randomly selected turkey weighs below 12 pounds. Your answer is . Round to 3 decimals and keep '0' before the decimal point.

Answer 1

Answer:

a. 0.443

b. 0.023

Step-by-step explanation:

Problems of normal distributions can be solved using 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.

The weight of turkeys is normally distributed with a mean of 22 pounds and a standard deviation of 5 pounds.

This means that $\mu = 22, \sigma = 5$

a. Find the probability that a randomly selected turkey weighs between 20 and 26 pounds.

This is the pvalue of Z when X = 26 subtracted by the pvalue of Z when X = 20. So

X = 26

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

$Z = \frac{26 - 22}{5}$

$Z = 0.8$

$Z = 0.8$ has a pvalue of 0.788

X = 20

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

$Z = \frac{20 - 22}{5}$

$Z = -0.4$

$Z = -0.4$ has a pvalue of 0.345

0.788 - 0.345 = 0.443

The answer is 0.443

b. Find the probability that a randomly selected turkey weighs below 12 pounds.

This is the pvalue of Z when X = 12. So

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

$Z = \frac{12 - 22}{5}$

$Z = -2$

$Z = -2$ has a pvalue of 0.023

The answer is 0.023.