5.1 (contd.) — Tinkering with the Median

5.1 (contd.) — Tinkering with the Median

We know that the median^? is the middle value in the sorted data — there are an equal number of values less than it and greater than it. Will including a new value to the data increase or decrease the median?

Let us consider the following data. The median of this data is 8: 3, 5, 7, 8, 9, 11, 13. If we insert a new value 11 (greater than 8), the sorted list becomes 3, 5, 7, 8, 9, 11, 11, 13 with 8 values — the two middle values are 8 and 9, median = \(\frac{8+9}{2}=8.5\). The median increased.

Suppose we include a value 11. The new value included is greater than the median — the median does not become larger by 8 as there are more values greater than it. Therefore, the median will also increase. The median, was 8 earlier, will be the average of the two middle values 8 and 11, which is 9.5.

We can similarly argue that when a value less than the median is included, the median will decrease.

Finding the Unknown

Coach Balwan's Kushti Players

Coach Balwan noted down the weights of the kushti players (wrestlers) and the mean as about 50 kg. For one value that was written down got smudged. Can you find out the missing value?

Average weight of the players = \(\frac{\text{Sum of weights}}{\text{Number of players}}\). Let the unknown weight be \(w\) kg. \(\frac{42+40+39+33+48+38+42+35+32+w}{10}=39.2\).

Simplifying: \(349+w=392\), so \(w=392-349=43\). The missing value is 43 kg.

Venkayya's Coconut Harvest

Venkayya keeps track of the coconut harvest in his farm. He calculates the average harvest per tree as 25.6. His son verifies the results and finds that one tree's harvest count is incorrectly noted as 3 more than the actual number. Can you find the correct average if the number of trees is 15?

Average harvest per tree = \(\frac{\text{Total number of coconuts harvested}}{\text{Number of trees}}\). Let the initial total be \(z\). Then \(\frac{z}{15}=25.6\) → \(z=384\). The true total = \(384-3=381\). Correct average = \(\frac{381}{15}=25.4\).

Mean and Median with Frequencies

A teacher asked how many family members each student has, and the data is shown in the table below. What is the average family size of this class?

Some of you may have thought, "Easy! It will be \(\frac{3+4+5+6+7+8+9+10}{8}=\frac{52}{8}=6.5\)." Remember that finding the average involves adding all the values in the data. The number 3 occurs three times; the number 4 occurs eleven times; and so on. Do these reflect in your calculation?

Accounting for the frequencies of each value, the average will be: \[\frac{(3\times 3)+(4\times 11)+(5\times 9)+(6\times 7)+(7\times 3)+(8\times 2)+(9\times 1)+(10\times 1)}{3+11+9+7+3+2+1+1}=\frac{9+44+45+42+21+16+9+10}{37}=\frac{196}{37}\approx 5.30.\]

Oh! That is close to \(5.22\) (a rounded version of the NCERT calculation). The average family size of this class is approximately 5.3.

What is the Median Family Size?

We know there are 36 values in the data. The median would be the average of the 18th and 19th values when the data is arranged in order.

Do we need to write all the 36 numbers in order to find this? No — we can use the frequency table. Successively add the frequencies starting from the smallest value until we reach 18 and 19. Adding the frequencies of 3 and 4, we get \(3+11=14\). Adding the frequencies of 3, 4 and 5, we get \(14+9=23\). Therefore, both the 18th and 19th values are 5, so the median is 5.

Spreadsheets — Sudhakar's Mid-Term Marks

Sudhakar has collected the mid-term exam marks obtained by his Grade 8 students in the following table (subjects: Odia, Telugu, English, Maths, Social Science, Science). He now wants to calculate the total marks scored by each student and the class average per subject.

Is there any way to further quicken this process? Spreadsheets^? — a digital grid of rows and columns made of cells^? — allow us to type in numbers and apply formulas that compute sums and averages automatically.

Cells are named and referred to by the column headers (A, B, C, ...) and row headers (1, 2, 3, ...). For instance, Farooq's score in Mathematics is in cell E5.

We can describe a row of cells by an expression of the form Start:End, indicating the first and last cell in the row. For instance, Nagesh's marks are described by the expression B3:G3; while Gowri's marks in Odia, Telugu and English are described by the expression B7:D7. We can also use similar expressions to describe columns. For instance D2:D6 describes the marks in English for the first five students.

We can also write a formula to compute the sum of a row or column. For instance =SUM(B3:G3) calculates the total marks for Nagesh across all subjects, while =AVERAGE(B7:D7) calculates Gowri's average marks across Odia, Telugu and English.

Find out if the class average marks in Odia is greater than the class average marks in Telugu. Show the average marks in other subjects after the last row by typing the appropriate formula. Get the total scores of each student by typing the appropriate formula.