Question

1. refer to the data set. 1 2 3 3 4 4 4 4 5 5 6 7 a. what happens to the mean and standard - deviation of the data set when the 7 is changed to a 70? b. for the data set with the value of 70, why would the median be a better choice for the measure of center than the mean?

Explanation:

Step1: Calculate the original mean

The original data - set is \(1,2,3,3,4,4,4,4,5,5,6,7\). The sum of the original data - set \(S_1=1 + 2+3+3+4+4+4+4+5+5+6+7=\sum_{i = 1}^{12}x_i=48\). The number of data points \(n = 12\). The original mean \(\bar{x}_1=\frac{S_1}{n}=\frac{48}{12}=4\).

Step2: Calculate the original standard - deviation

First, calculate the squared differences \((x_i-\bar{x}_1)^2\) for each \(x_i\). Then the variance \(s_1^2=\frac{\sum_{i = 1}^{n}(x_i - \bar{x}_1)^2}{n - 1}\).
\(\sum_{i = 1}^{12}(x_i - 4)^2=(1 - 4)^2+(2 - 4)^2+(3 - 4)^2+(3 - 4)^2+(4 - 4)^2+(4 - 4)^2+(4 - 4)^2+(4 - 4)^2+(5 - 4)^2+(5 - 4)^2+(6 - 4)^2+(7 - 4)^2\)
\(=9 + 4+1 + 1+0+0+0+0+1+1+4+9 = 30\). The original variance \(s_1^2=\frac{30}{11}\approx2.73\), and the original standard - deviation \(s_1=\sqrt{\frac{30}{11}}\approx1.65\).

Step3: Calculate the new mean

When the \(7\) is changed to \(70\), the new sum \(S_2=48-7 + 70=111\). The new mean \(\bar{x}_2=\frac{S_2}{n}=\frac{111}{12}=9.25\).

Step4: Calculate the new standard - deviation

First, calculate the squared differences \((x_i-\bar{x}_2)^2\) for each \(x_i\) in the new data - set. The new variance \(s_2^2=\frac{\sum_{i = 1}^{n}(x_i - \bar{x}_2)^2}{n - 1}\).
The new sum of squared differences \(\sum_{i = 1}^{12}(x_i - 9.25)^2=(1 - 9.25)^2+(2 - 9.25)^2+(3 - 9.25)^2+(3 - 9.25)^2+(4 - 9.25)^2+(4 - 9.25)^2+(4 - 9.25)^2+(4 - 9.25)^2+(5 - 9.25)^2+(5 - 9.25)^2+(6 - 9.25)^2+(70 - 9.25)^2\)
\(=(- 8.25)^2+(-7.25)^2+(-6.25)^2+(-6.25)^2+(-5.25)^2+(-5.25)^2+(-5.25)^2+(-5.25)^2+(-4.25)^2+(-4.25)^2+(-3.25)^2+(60.75)^2\)
\(=68.0625+52.5625+39.0625+39.0625+27.5625+27.5625+27.5625+27.5625+18.0625+18.0625+10.5625 + 3690.5625=3946.25\). The new variance \(s_2^2=\frac{3946.25}{11}\approx358.75\), and the new standard - deviation \(s_2=\sqrt{\frac{3946.25}{11}}\approx18.94\).
So, the mean increases from \(4\) to \(9.25\) and the standard deviation increases from approximately \(1.65\) to approximately \(18.94\).

Step5: Explain why the median is a better measure of center

The original data - set has \(n = 12\) data points. The median of the original data - set (and the new data - set since the number of data points is the same) is the average of the \(6^{th}\) and \(7^{th}\) ordered data points. For the original and new data - sets, the ordered data points are considered. The median is \(\frac{4 + 4}{2}=4\). The mean is affected by the outlier (the value of \(70\) in the new data - set), while the median is not. The median represents the middle value of the data - set and is not influenced by extreme values. So, when there is an outlier, the median is a better measure of the center of the data - set.

Answer:

a. The mean increases from \(4\) to \(9.25\) and the standard deviation increases from approximately \(1.65\) to approximately \(18.94\).
b. The median is not affected by the outlier (\(70\)), while the mean is. The median represents the middle - value of the data - set and is a more robust measure of center in the presence of outliers.