Question

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.

$\sigma^{2}=\frac{(87 - 78)^{2}+(46 - 78)^{2}+(90 - 78)^{2}+(78 - 78)^{2}+(89 - 78)^{2}}{5}$

$\sigma^{2}=\frac{(9)^{2}-(32)^{2}+(12)^{2}+0^{2}+(11)^{2}}{5}$

$\sigma^{2}=\frac{81 - 1024+144+0+121}{5}$

$\sigma^{2}=\frac{-678}{5}=-135.6$

what is the first error cara made in computing the variance?

Explanation:

Step1: Recall the formula for variance

The formula for variance of a population is \(\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\mu)^{2}}{n}\), where \(x_{i}\) are the data points, \(\mu\) is the mean, and \(n\) is the number of data points. When expanding \((a - b)^{2}=a^{2}-2ab + b^{2}\), for \((46 - 78)^{2}\), it should be \((-32)^{2}=(-32)\times(-32)=1024\), not \(-(32)^{2}\).

Answer:

Cara made a mistake in squaring \((46 - 78)\). It should be \((- 32)^{2}=1024\) instead of \(-(32)^{2}\) in the second step of her calculation.