Question

Big chickens: The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1511 grams and standard deviation 198 grams. Use the TI-84 Plus calculator to answer the following. (a) What proportion of broilers weigh between 1143 and 1242 grams?

Answer 1

Answer:

5.55% of broilers weigh between 1143 and 1242 grams

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 1511, \sigma = 198$

What proportion of broilers weigh between 1143 and 1242 grams?

This is the pvalue of Z when X = 1242 subtracted by the pvalue of Z when X = 1143.

X = 1242

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

$Z = \frac{1242 - 1511}{198}$

$Z = -1.36$

$Z = -1.36$ has a pvalue of 0.0869

X = 1143

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

$Z = \frac{1143 - 1511}{198}$

$Z = -1.36$

$Z = -1.86$ has a pvalue of 0.0314

0.0869 - 0.0314 = 0.0555

5.55% of broilers weigh between 1143 and 1242 grams