cara computes the mean and variance for the set 87, 46, 90, 78, and 89. she finds the mean to be 78. her…

cara computes the mean and variance for the set 87, 46, 90, 78, and 89. she finds the mean to be 78. her steps for finding the variance are shown below.\n$sigma^{2}=\frac{(87 - 78)^{2}+(46 - 78)^{2}+(90 - 78)^{2}+(78 - 78)^{2}+(89 - 78)^{2}}{5}$\n$sigma^{2}=\frac{(9)^{2}-(32)^{2}+(12)^{2}+0^{2}+(11)^{2}}{5}$\n$sigma^{2}=\frac{81 - 1024+144 + 0+121}{5}$\n$sigma^{2}=\frac{-678}{5}=-135.6$\nwhat is the first error cara made in computing the variance?

cara computes the mean and variance for the set 87, 46, 90, 78, and 89. she finds the mean to be 78. her steps for finding the variance are shown below.\n$sigma^{2}=\frac{(87 - 78)^{2}+(46 - 78)^{2}+(90 - 78)^{2}+(78 - 78)^{2}+(89 - 78)^{2}}{5}$\n$sigma^{2}=\frac{(9)^{2}-(32)^{2}+(12)^{2}+0^{2}+(11)^{2}}{5}$\n$sigma^{2}=\frac{81 - 1024+144 + 0+121}{5}$\n$sigma^{2}=\frac{-678}{5}=-135.6$\nwhat is the first error cara made in computing the variance?

Answer

Explanation:

Step1: Recall variance formula

The formula for the variance $\sigma^{2}$ of a sample of data points $x_1,x_2,\cdots,x_n$ with mean $\bar{x}$ is $\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n}$. When squaring the differences $(x_i - \bar{x})$, we should get non - negative values.

Step2: Analyze Cara's second step

In the step $\sigma^{2}=\frac{(9)^{2}-(32)^{2}+(12)^{2}+0^{2}+(11)^{2}}{5}$, Cara made an error. The correct expansion of $(46 - 78)^2$ is $( - 32)^2=32^{2}$, not $-(32)^{2}$. When we square a number, the result is always non - negative.

Answer:

Cara incorrectly wrote $(46 - 78)^2$ as $-(32)^{2}$ instead of $(32)^{2}$ in the second step of computing the variance.