Question 15 Marks
The following table shows the weight of 12 players:
Find the median and mean weights.
Using empirical formula, calculate its mode.
| Weight (in kg) | 48 | 50 | 52 | 54 | 58 |
| Number of players | 4 | 3 | 2 | 2 | 1 |
Using empirical formula, calculate its mode.
Answer
View full question & answer→We prepare the table as given below:
Number of terms (N) is 12, which is an even number.
$\text{Median}=\frac{1}{2}\Big\{\Big(\frac{\text{N}}{2}\Big)\text{th observation}+\Big(\frac{\text{n}}{2}+1\Big)\text{th observation}\Big\}$
$=\{6\ \text{th observation}+7\text{th observation}\}$
$=\frac{1}{2}\{50+50\}$
$\text{Median}=50$
$\text{Mean}=\frac{\sum(\text{f}_\text{i}\times\text{x}_\text{i})}{\sum\text{f}_\text{i}}$
$=\frac{612}{12}$
Mean = 51
Using empirical formula:
Mode = 3(Median) - 2(Mean)
= 150 - 102
Mode = 48
Hence, the median is 50, the mean is 51 and the mode is 48.
| Weight (in kg) (x) | Number of players (f) | c.f | x × f |
| 48 | 4 | 4 | 192 |
| 50 | 3 | 7 | 150 |
| 52 | 2 | 9 | 104 |
| 54 | 2 | 11 | 108 |
| 58 | 1 | 12 | 58 |
| Total | 12 | 612 |
$\text{Median}=\frac{1}{2}\Big\{\Big(\frac{\text{N}}{2}\Big)\text{th observation}+\Big(\frac{\text{n}}{2}+1\Big)\text{th observation}\Big\}$
$=\{6\ \text{th observation}+7\text{th observation}\}$
$=\frac{1}{2}\{50+50\}$
$\text{Median}=50$
$\text{Mean}=\frac{\sum(\text{f}_\text{i}\times\text{x}_\text{i})}{\sum\text{f}_\text{i}}$
$=\frac{612}{12}$
Mean = 51
Using empirical formula:
Mode = 3(Median) - 2(Mean)
= 150 - 102
Mode = 48
Hence, the median is 50, the mean is 51 and the mode is 48.