Understanding the Frequency Distribution Problem

The problem provides a frequency distribution of data grouped into classes. We are given the classes, some frequencies directly, and other frequencies in terms of variables 'p' and 'q'. We are also given the total frequency and the mean of this distribution. The question asks how the mean changes if the frequency of each class is doubled.

Analyzing the Given Frequency Distribution Data

Here is the frequency distribution provided:

We are given that the total frequency, $\sum f_i$, is 120, and the mean, $\bar{x}$, is 50.

Effect of Doubling Frequencies on the Mean

The mean of a frequency distribution is calculated using the formula:

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$

where $f_i$ is the frequency of each class and $x_i$ is the midpoint of each class. $\sum f_i$ is the total frequency, often denoted by $N$.

The question asks what happens if the frequency of each class is doubled. Let the new frequencies be $f'_i$. According to the question, $f'_i = 2 f_i$ for every class.

Let's calculate the new total frequency, $\sum f'_i$:

$\sum f'_i = \sum (2 f_i)$

Using the property of summation, we can take the constant factor (2) outside the summation:

$\sum f'_i = 2 \sum f_i$

So, the new total frequency is double the original total frequency. The original total frequency was 120, so the new total frequency is $2 \times 120 = 240$.

Now, let's calculate the new mean, $\bar{x}'$, using the formula with the new frequencies $f'_i$ and the same midpoints $x_i$:

$\bar{x}' = \frac{\sum f'_i x_i}{\sum f'_i}$

Substitute $f'_i = 2 f_i$ and $\sum f'_i = 2 \sum f_i$:

$\bar{x}' = \frac{\sum (2 f_i) x_i}{2 \sum f_i}$

Again, we can take the constant factor (2) outside the summation in the numerator:

$\bar{x}' = \frac{2 \sum f_i x_i}{2 \sum f_i}$

We can see that the factor of 2 in the numerator and the denominator cancels out:

$\bar{x}' = \frac{\sum f_i x_i}{\sum f_i}$

This is the exact same formula as the original mean, $\bar{x}$.

Conclusion on Mean with Doubled Frequencies

The calculation shows that if the frequency of every class in a frequency distribution is multiplied by the same constant factor (in this case, 2), the mean of the distribution does not change. The scaling factor cancels out in the mean formula.

Therefore, if the frequency of each class is doubled, the mean would be the same as the original mean.

Given that the original mean is 50, the new mean would also be 50.

Calculation of Unknown Frequencies (Optional)

Although not necessary to answer the specific question about the mean change, we can determine the values of p and q using the given total frequency and mean.

Sum of frequencies: $17 + (p+q) + 32 + (p-3q) + 19 = 120$

$68 + 2p - 2q = 120$

$2p - 2q = 52$

$p - q = 26$ (Equation 1)

Mean calculation: $\frac{(17 \times 10) + ((p+q) \times 30) + (32 \times 50) + ((p-3q) \times 70) + (19 \times 90)}{120} = 50$

$\frac{170 + 30p + 30q + 1600 + 70p - 210q + 1710}{120} = 50$

$3480 + 100p - 180q = 6000$

$100p - 180q = 2520$

Dividing by 20: $5p - 9q = 126$ (Equation 2)

Solving the system of equations:

• From Equation 1, $p = q + 26$.
• Substitute into Equation 2: $5(q + 26) - 9q = 126$
• $5q + 130 - 9q = 126$
• $-4q = -4 \implies q = 1$
• Substitute $q=1$ into $p = q+26$: $p = 1 + 26 = 27$.

So, $p=27$ and $q=1$. The frequencies are $17, 28, 32, 24, 19$. Summing these confirms the total frequency is 120. Calculating the mean with these frequencies confirms it is 50. However, these values do not change the principle that scaling all frequencies by a constant factor leaves the mean unchanged.

Revision Table: Frequency Distribution Key Concepts

Additional Information: Properties of Mean

The mean is a widely used measure of central tendency. It has several important properties:

• It is sensitive to extreme values (outliers).
• The sum of deviations of observations from the mean is always zero ($\sum (x_i - \bar{x}) = 0$).
• If each observation in a dataset is increased or decreased by a constant, the mean is also increased or decreased by the same constant.
• If each observation in a dataset is multiplied or divided by a constant, the mean is also multiplied or divided by the same constant.
• In a frequency distribution, multiplying all frequencies by a constant factor scales the total frequency and the sum of $(f_i \times x_i)$ by the same factor. This factor cancels out in the mean calculation, leaving the mean unchanged. This is the property demonstrated in the problem.