Question

1 2 3 3 4 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?
when the data set 7 changed to a 70, the
mean and the standard deviation increases
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?
the median would be a better choice for

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 data \(S_1=1 + 2+3+3+4+4+4+4+5+5+6+7=48\).
The number of data points \(n = 12\).
The original mean \(\bar{x}_1=\frac{S_1}{n}=\frac{48}{12} = 4\).

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

Step2: Calculate the original standard deviation

The formula for the standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n}}\).
For the original data:
\(\sum_{i = 1}^{12}(x_i - 4)^2=(1 - 4)^2+(2 - 4)^2+(3 - 4)^2+(3 - 4)^2+(4 - 4)^2+(4 - 4)^2+(4 - 4)^2+(4 - 4)^2+(5 - 4)^2+(5 - 4)^2+(6 - 4)^2+(7 - 4)^2\)
\(=9 + 4+1+1+0+0+0+0+1+1+4+9=30\).
The original standard deviation \(s_1=\sqrt{\frac{30}{12}}\approx1.58\).

For the new data:
\(\sum_{i = 1}^{12}(x_i - 9.25)^2=(1 - 9.25)^2+(2 - 9.25)^2+(3 - 9.25)^2+(3 - 9.25)^2+(4 - 9.25)^2+(4 - 9.25)^2+(4 - 9.25)^2+(4 - 9.25)^2+(5 - 9.25)^2+(5 - 9.25)^2+(6 - 9.25)^2+(70 - 9.25)^2\)
\(=(- 8.25)^2+(-7.25)^2+(-6.25)^2+(-6.25)^2+(-5.25)^2+(-5.25)^2+(-5.25)^2+(-5.25)^2+(-4.25)^2+(-4.25)^2+(-3.25)^2+(60.75)^2\)
\(=68.0625+52.5625+39.0625+39.0625+27.5625+27.5625+27.5625+27.5625+18.0625+18.0625+10.5625+3690.5625\)
\(=4146.375\).
The new standard deviation \(s_2=\sqrt{\frac{4146.375}{12}}\approx18.56\).

Step3: Explain why median is a better measure

The median is the middle - value of a data set. For the original data set \(1,2,3,3,4,4,4,4,5,5,6,7\), the median (average of the 6th and 7th values) is \(\frac{4 + 4}{2}=4\).
When \(7\) is changed to \(70\), the new data set is \(1,2,3,3,4,4,4,4,5,5,6,70\), and the median (average of the 6th and 7th values) is still \(\frac{4+4}{2} = 4\).
The mean is affected by extreme values (outliers). Since \(70\) is an outlier, the median is more robust (less affected by outliers) than the mean as a measure of central tendency.

Answer:

a. The mean increases from \(4\) to \(9.25\) and the standard deviation increases from approximately \(1.58\) to approximately \(18.56\).
b. The median is a better measure of center than the mean because the median is not affected by extreme values (outliers), while the mean is. In this case, changing \(7\) to \(70\) (an outlier) changes the mean but not the median.