Question

following results.
mean: 76 median: 77 range: 29 iqr: 12
the test was interrupted early due to unannounced fire drill, so mrs. lloyd decided to give each student present that day an extra 5 points on the test.
what are the mean, median, range, and iqr of the data set after 5 points is added to each students test?

• press each hotspot.
• label the corresponding number below with the requested value.

mean: 1 median: 2 range: 3 iqr: 4

Explanation:

Step1: Effect on mean

If we add a constant \( c = 5 \) to each data - point, the mean of the data set changes. The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\). If \(y_{i}=x_{i}+5\), then \(\bar{y}=\frac{\sum_{i = 1}^{n}(x_{i}+5)}{n}=\frac{\sum_{i = 1}^{n}x_{i}+5n}{n}=\frac{\sum_{i = 1}^{n}x_{i}}{n}+5\). Given the original mean \(\bar{x}=76\), the new mean is \(76 + 5=81\).

Step2: Effect on median

The median is the middle value of a sorted data set. If we add a constant \(c = 5\) to each data - point, the position of the middle value remains the same, but its value increases by \(5\). Given the original median \(M = 77\), the new median is \(77+5 = 82\).

Step3: Effect on range

The range is \(R=\text{Max}-\text{Min}\). If \(y_{i}=x_{i}+5\), then \(R_y=\text{Max}(y_{i})-\text{Min}(y_{i})=(\text{Max}(x_{i}) + 5)-(\text{Min}(x_{i})+5)=\text{Max}(x_{i})-\text{Min}(x_{i})\). Given the original range \(R = 29\), the new range is \(29\).

Step4: Effect on IQR

The inter - quartile range \(IQR = Q_{3}-Q_{1}\). If \(y_{i}=x_{i}+5\), then \(Q_{3y}=Q_{3x}+5\) and \(Q_{1y}=Q_{1x}+5\). So \(IQR_y=(Q_{3x}+5)-(Q_{1x}+5)=Q_{3x}-Q_{1x}\). Given the original \(IQR = 12\), the new \(IQR\) is \(12\).

Answer:

1. \(81\)
2. \(82\)
3. \(29\)
4. \(12\)