Marks 13 33 51 76 83 No. of Students 53 13 24 23 16
The mean of a frequency distribution represents the average value of the data. It is calculated using the formula:
$ \text{Mean} (\bar{x}) = \frac{\sum (f \cdot x)}{\sum f} $
In this formula, '$f$' represents the frequency of each data point (here, 'No. of Students'), and '$x$' represents the value of the data point (here, 'Marks'). The term '$ \sum (f \cdot x) $' signifies the sum of the products of each frequency and its corresponding value, while '$ \sum f $' represents the total sum of all frequencies.
The data provided in the question is as follows:
Note on Data Consistency: It's important to notice that there are 7 distinct 'Marks' values listed, but there are 10 'No. of Students' (frequency) values. For a standard frequency distribution calculation, the number of values should match the number of frequencies. We will proceed by assuming that the first 7 frequency values correspond directly to the 7 listed marks, as this is a common approach when encountering such discrepancies.
1. Organize the Data: We'll create a table to list the Marks (x), the corresponding No. of Students (f), and calculate the product (f * x) for each pair.
| Marks (x) | No. of Students (f) | Product (f * x) |
|---|---|---|
| 13 | 5 | 65 |
| 33 | 3 | 99 |
| 35 | 1 | 35 |
| 17 | 3 | 51 |
| 68 | 2 | 136 |
| 3 | 4 | 12 |
| 16 | 2 | 32 |
| Total | 20 | 430 |
2. Sum of Frequencies ($ \sum f $): Add up the frequencies from the table.
$ \sum f = 5 + 3 + 1 + 3 + 2 + 4 + 2 = 20 $
3. Sum of Products ($ \sum (f \cdot x) $): Add up the values in the 'Product (f * x)' column.
$ \sum (f \cdot x) = 65 + 99 + 35 + 51 + 136 + 12 + 32 = 430 $
4. Calculate the Mean: Apply the mean formula using the sums calculated.
$ \bar{x} = \frac{\sum (f \cdot x)}{\sum f} = \frac{430}{20} $
$ \bar{x} = 21.5 $
Based on the standard calculation method applied to the provided data (and interpreting the frequency list as described), the calculated mean for this distribution is 21.5. This value is derived directly from the marks and the corresponding frequencies.
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 |