Question

Hens usually begin laying eggs when they are about 6 months old. Young hens tend to lay smaller eggs, often weighing less than the desired minimum weight of 54 grams. a) Theaverageweightoftheeggsproducedbythe young hens is 50.9 grams, and only 28% of their eggs exceed the desired minimum weight. If a Normal model is appropriate, what would the standard devia- tion of the egg weights be? b) Bythetimethesehenshavereachedtheageof1year, the eggs they produce average 67.1 grams, and 98% of them are above the minimum weight. What is the standard deviation for the appropriate Normal model for these older hens?

Answer 1

Answer:

a) The standard deviation of the egg's weights will be 3.37 grams.

b) The standard deviation is 4.55g.

Step-by-step explanation:

Normal model problems can be solved by the zscore formula.

On a normaly distributed set with mean \mu and standard deviation \sigma, the z-score of a value X is given by:

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

Each z-score value has an equivalent p-value, that represents the percentile that the value X is.

The desired minimum weight is 54 grams.

a) The average weight of the eggs produced by the young hens is 50.9 grams, and only 28% of their eggs exceed the desired minimum weight. If a Normal model is appropriate, what would the standard devia- tion of the egg weights be?

This means that a value of $X = 54$ is in the 100-28 = 72nd percentile.

What is the z-score of the p-value of 0.72? We have to look at the zscore table. This value is z = 0.92. This means that when $X = 54, Z = 0.92$. We also have that $\mu = 50.9$. So, we can apply the formula and find $\sigma$

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

$0.92 = \frac{54 - 50.9}{\sigma}$

$0.92\sigma = 3.1$

$\sigma = \frac{3.1}{0.92}$

$\sigma = 3.37$

The standard deviation of the egg's weights will be 3.37 grams.

b) By the time these hens have reached the age of 1 year, the eggs they produce average 67.1 grams, and 98% of them are above the minimum weight.

This means that a value of $X = 54$ is in the 2nd percentile.

What zscore has a pvalue of 0.02? This is a zscore of -2.88. So, when $X = 54, Z = -2.88$. We also have that $\mu = 67.1$. So, again, we can apply the formula and find $\sigma$.

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

$-2.88 = \frac{54 - 67.1}{\sigma}$

$-2.88\sigma = -13.1$*(-1)

$2.88\sigma = 13.1$

$\sigma = \frac{13.1}{2.88}$

$\sigma = 4.55$

The standard deviation is 4.55g.