Question

data were collected from a typing class with 45 students. the class had an average of 56 words per minute (wpm) with a standard deviation of 7 wpm. the teacher realized that the data from one student who typed 55 wpm was not entered. if this students results were included in the data set, what effect would it have on the standard deviation?
the standard deviation would be noticeably larger.
the standard deviation would be noticeably smaller.
the standard deviation would be about the same, perhaps a little larger.
the standard deviation would be about the same, perhaps a little smaller.

Explanation:

Step1: Analyze the relationship between the new data point and the mean

The original mean is \(56\) WPM. The new data point is \(55\) WPM. The difference between the new data point and the mean is \(|55 - 56|=1\).

Step2: Consider the impact on standard deviation formula

The formula for standard deviation is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\). When adding a new data point \(x_{new}\), we calculate \((x_{new}-\bar{x})^{2}\). Since \(1^{2}=1\), and the original standard - deviation is calculated based on \(\sum_{i = 1}^{45}(x_{i}-56)^{2}\) with \(n = 45\). The original sum of squares \(\sum_{i = 1}^{45}(x_{i}-56)^{2}=(n - 1)s^{2}=(45 - 1)\times7^{2}=44\times49 = 2156\). After adding the new data point, the new sum of squares is \(2156+1=2157\) and \(n = 46\). The new standard deviation \(s_{new}=\sqrt{\frac{2157}{45}}\approx\sqrt{47.93}\approx6.92\)

Answer:

The standard deviation would be about the same, perhaps a little smaller.