Marks 14 30 44 61 98 No. of Students 49 21 75 39 99
To find the mean of a frequency distribution, we calculate the sum of the products of each value and its corresponding frequency, and then divide by the total number of observations (total frequency).
The formula used is:
$ \text{Mean} (\bar{x}) = \frac{\sum (f \times x)}{\sum f} $
Where:
Let's set up a table to calculate '$f \times x$' for each row and find the sums:
| Marks ($x$) | No. of Students ($f$) | Product ($f \times x$) |
|---|---|---|
| 14 | 4 | $14 \times 4 = 56$ |
| 30 | 9 | $30 \times 9 = 270$ |
| 44 | 21 | $44 \times 21 = 924$ |
| 61 | 75 | $61 \times 75 = 4575$ |
| 98 | 39 | $98 \times 39 = 3822$ |
| Total | $\sum f = 4 + 9 + 21 + 75 + 39 = 148$ | $\sum (f \times x) = 56 + 270 + 924 + 4575 + 3822 = 9647$ |
Now, we plug the total sum of products and the total frequency into the mean formula:
$ \text{Mean} = \frac{\sum (f \times x)}{\sum f} $
Substitute the calculated values:
$ \text{Mean} = \frac{9647}{148} $
Performing the division gives the mean:
$ \text{Mean} \approx 65.18 $
This detailed calculation demonstrates the step-by-step process for finding the mean of the provided distribution.
Let $X_1, X_2, X_3$ be a random sample of size 3 from an absolutely continuous distribution that is symmetric about 0. For $i=1,2,3$, let $R_i$ denote the rank of $|X_i|$ among $|X_1|, |X_2|$ and $|X_3|$.
If $T^+ = \sum_{i=1, X_i>0}^3 R_i$
is the Willcoxon signed-rank statistic, then which of the following statements are true?,
The mean marks of the following distribution is:
| Marks Obtained | No. of Students |
| 81 | 15 |
| 35 | 4 |
| 73 | 3 |
| 56 | 16 |