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: Calculate each squared term

For \((87 - 78)^2\), \(87-78 = 9\), so \((87 - 78)^2=9^2 = 81\).
For \((46 - 78)^2\), \(46-78=-32\), so \((46 - 78)^2=(-32)^2 = 1024\) (not \(-(32)^2\)).
For \((90 - 78)^2\), \(90 - 78=12\), so \((90 - 78)^2=12^2 = 144\).
For \((78 - 78)^2\), \(78-78 = 0\), so \((78 - 78)^2=0^2 = 0\).
For \((89 - 78)^2\), \(89-78 = 11\), so \((89 - 78)^2=11^2 = 121\).

Step2: Identify the error

The formula for variance is \(\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\mu)^{2}}{n}\), where \((x_{i}-\mu)^{2}\) is always non - negative. In Cara's second step, she wrote \((46 - 78)^2\) as \(-(32)^2\) instead of \((-32)^2\). The negative sign before \((32)^2\) is incorrect.

Answer:

Cara incorrectly wrote \((46 - 78)^2\) as \(-(32)^2\) instead of \((-32)^2\) in the second step.