Question

calculate the score for jackie joyner - kersees jump when it is measured in feet. what do you notice? a local farmer sells chicken eggs at a farmers market every weekend. she keeps track of the number of dozens of eggs sold each weekend for an entire year, and finds the mean is 14.7 dozen and the standard deviation is 3.0 dozen. what is the variance for the distribution of dozens of eggs sold? suppose the farmer converts all the values from dozens of eggs to number of eggs. what is the new mean and new standard deviation? what is the variance for the distribution of number of eggs sold?

Explanation:

Step1: Recall variance - standard deviation relationship

The variance $\sigma^{2}$ is the square of the standard deviation $\sigma$. Given $\sigma = 3.0$ dozen for the distribution of dozens of eggs sold.
$\sigma^{2}=(3.0)^{2}$

Step2: Calculate the variance

$\sigma^{2}=9.0$ dozen$^{2}$

Step3: Convert mean and standard - deviation to number of eggs

Since 1 dozen = 12 eggs, if the mean in dozens is $\mu_{d}=14.7$ dozen and standard - deviation in dozens is $\sigma_{d}=3.0$ dozen.
The new mean $\mu_{e}$ (in number of eggs) is $\mu_{e}=14.7\times12 = 176.4$ eggs.
The new standard deviation $\sigma_{e}$ (in number of eggs) is $\sigma_{e}=3.0\times12=36$ eggs.

Step4: Calculate the variance for number of eggs

The variance for number of eggs $\sigma_{e}^{2}$ is the square of the standard deviation of number of eggs. Since $\sigma_{e} = 36$ eggs, $\sigma_{e}^{2}=(36)^{2}=1296$ eggs$^{2}$

Answer:

The variance for the distribution of dozens of eggs sold is 9.0 dozen$^{2}$.
The new mean in number of eggs is 176.4 eggs and the new standard deviation is 36 eggs.
The variance for the distribution of number of eggs sold is 1296 eggs$^{2}$.