Question

from unit 1, lesson 12
here is a data set:
5
10
10
10
15
100
a. after studying the data, the researcher realized that the value 100 was meant to be recorded as 15. what happens to the mean and standard deviation of the data set when the 100 is changed to a 15?

Explanation:

Step1: Calculate the original mean

The original data set is \(5, 10, 10, 10, 15, 100\). The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), where \(n = 6\) and \(\sum_{i=1}^{6}x_{i}=5 + 10+10+10+15+100=150\). So the original mean \(\bar{x}_{1}=\frac{150}{6}=25\).

Step2: Calculate the new mean

The new data - set after changing \(100\) to \(15\) is \(5, 10, 10, 10, 15, 15\). Now \(\sum_{i = 1}^{6}x_{i}=5 + 10+10+10+15+15=65\), and the new mean \(\bar{x}_{2}=\frac{65}{6}\approx10.83\). The mean decreases.

Step3: Calculate the original standard deviation

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\).
For the original data set:
\((5 - 25)^{2}=400\), \((10 - 25)^{2}=225\), \((10 - 25)^{2}=225\), \((10 - 25)^{2}=225\), \((15 - 25)^{2}=100\), \((100 - 25)^{2}=5625\).
\(\sum_{i = 1}^{6}(x_{i}-25)^{2}=400+225+225+225+100+5625=6800\).
The original standard deviation \(s_{1}=\sqrt{\frac{6800}{5}}=\sqrt{1360}\approx36.88\).

Step4: Calculate the new standard deviation

For the new data set:
\((5 - 10.83)^{2}\approx34.0\), \((10 - 10.83)^{2}\approx0.69\), \((10 - 10.83)^{2}\approx0.69\), \((10 - 10.83)^{2}\approx0.69\), \((15 - 10.83)^{2}\approx17.39\), \((15 - 10.83)^{2}\approx17.39\).
\(\sum_{i = 1}^{6}(x_{i}-10.83)^{2}\approx34.0+0.69+0.69+0.69+17.39+17.39 = 70.85\).
The new standard deviation \(s_{2}=\sqrt{\frac{70.85}{5}}=\sqrt{14.17}\approx3.76\). The standard deviation decreases.

Answer:

The mean and the standard deviation of the data - set decrease.