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practice 4 (from unit 1, lesson 9) a student has these scores on ten as…

Question

practice 4 (from unit 1, lesson 9)
a student has these scores on ten assignments.
0 40 60 70 75 80 85 95 95 100
the teacher is considering dropping a lowest score.
what effect does eliminating the lowest value, 0, from the data set have on the mean and median?

Explanation:

Step1: Calculate the original mean

The original data - set is \(0,40,60,70,75,80,85,95,95,100\). The sum of the data - set \(S_1=0 + 40+60 + 70+75+80+85+95+95+100 = 705\). The number of data points \(n_1 = 10\). The original mean \(\bar{x}_1=\frac{S_1}{n_1}=\frac{705}{10}=70.5\).

Step2: Calculate the new mean

After dropping the lowest score \(0\), the new data - set is \(40,60,70,75,80,85,95,95,100\). The sum of the new data - set \(S_2=40 + 60+70+75+80+85+95+95+100 = 700\). The number of data points \(n_2 = 9\). The new mean \(\bar{x}_2=\frac{S_2}{n_2}=\frac{700}{9}\approx77.78\). So the mean increases.

Step3: Calculate the original median

For a data - set with \(n = 10\) (an even number of data points), the median is the average of the \(\frac{n}{2}\)th and \((\frac{n}{2}+1)\)th ordered values. \(\frac{n}{2}=5\) and \(\frac{n}{2}+1 = 6\). The original median \(M_1=\frac{75 + 80}{2}=77.5\).

Step4: Calculate the new median

For a data - set with \(n = 9\) (an odd number of data points), the median is the \(\frac{n + 1}{2}\)th ordered value. \(\frac{n+1}{2}=5\). The new median \(M_2 = 80\). So the median increases.

Answer:

Both the mean and the median increase.