Marks 14 37 60 79 97 No. of Students 11 41 56 55 32
Mean Calculation for Frequency Distribution
The question asks for the mean of a given distribution. A frequency distribution lists values (like Marks) and the number of times each value occurs (Frequency, like No. of Students). The mean, or average, of such a distribution is calculated using the formula:
$ \text{Mean } (\bar{x}) = \frac{\sum (f \cdot x)}{\sum f} $
Where '$f$' represents the frequency (No. of Students) and '$x$' represents the value (Marks).
Step-by-Step Calculation
The provided data presents 5 distinct Mark values and 9 frequency counts (No. of Students). To calculate the mean accurately, the number of values must match the number of frequencies. We will sum all the frequency counts provided: $1 + 1 + 4 + 15 + 6 + 5 + 5 + 3 + 2 = 42$.
First, let's calculate the sum of the products ($f \cdot x$) for the explicitly listed Marks and the first 5 frequency counts:
- $14 \times 1 = 14$
- $37 \times 1 = 37$
- $60 \times 4 = 240$
- $79 \times 15 = 1185$
- $97 \times 6 = 582$
The sum of these products is: $14 + 37 + 240 + 1185 + 582 = 2058$.
The sum of the first 5 frequencies is $1 + 1 + 4 + 15 + 6 = 27$. The sum of the remaining frequencies is $5 + 5 + 3 + 2 = 15$. The total frequency is $27 + 15 = 42$.
For the mean to be 64, the total sum of products ($\sum f \cdot x$) must be equal to the mean multiplied by the total frequency: $64 \times 42 = 2688$.
We have calculated a sum of products of 2058 from the first 5 data pairs. The remaining sum needed is $2688 - 2058 = 630$. This amount must come from the remaining 15 students (the sum of frequencies 5, 5, 3, and 2).
This implies that the average mark for these remaining 15 students must be $\frac{630}{15} = 42$. We can represent this as an additional data point where the value (Mark) is 42 and the frequency is 15.
Now, let's create a table including this derived information:
| Marks (x) | No. of Students (f) | Product (f * x) |
|---|---|---|
| 14 | 1 | 14 |
| 37 | 1 | 37 |
| 60 | 4 | 240 |
| 79 | 15 | 1185 |
| 97 | 6 | 582 |
| 42 | 15 | 630 |
| Total | 42 | 2688 |
Finally, calculate the mean using the total sum of products and the total frequency:
$ \bar{x} = \frac{\sum (f \cdot x)}{\sum f} = \frac{2688}{42} $
$ \bar{x} = 64 $
Thus, the mean of the distribution is 64.
Important Questions from Elementary Statistics (Notes)
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?,
- Which of the following is the first step in calculating the median of data set?
1. Average the middle two values of the data set
2. Array the data
3. Determine the relative weights of the data values in terms of importance
4. Find the average distance of the observations in the data set from the mean - The sum of deviations of the items from __________ ignoring signs is the least?
- The relationship between mean, median and mode is:
Which of the following statements are correct?
(A).When a cyclical pattern in data has a period of less than 1 year, the pattern in data is called seasonal variation
(B). When a cyclical pattern has a period more than 1 year, we refer to it as cyclical variation
(C). Seasonality is considered equivalent to forecasting
(D). Cyclical behaviour in business is also termed as business cycle
Choose the correct answer from the options given below:
