Consider the following frequency distribution table: Class Interval 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Frequency If the total frequency is and the median is , then find the value of and .

Explanation: Solution 1. Equation from Total Frequency Summing all frequencies to : 2. Cumulative Frequency (CF) Table Class Interval Frequency (f) Cumulative Frequency (cf) 3. Identifying Median Class Since the median is , the Median Class is . (CF of the class before the median class) 4. Applying the Median Formula Since frequency must be an integer, we round to . 5. Finding Using Eq. 1 : Final Answer The values are and .

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NIMCET 2021 — Mathematics PYQ

NIMCET | Mathematics | 2021

Consider the following frequency distribution table:

Class Interval	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Frequency	$180$	$f_1$	$34$	$180$	$136$	$f_2$	$50$

If the total frequency is $686$ and the median is $42.6$ , then find the value of $f_{1}$ and $f_{2}$ .

Explanation

Solution

1. Equation from Total Frequency

Summing all frequencies to $686$ :

$180 + f_1 + 34 + 180 + 136 + f_2 + 50 = 686$

$580 + f_1 + f_2 = 686$

$f_1 + f_2 = 686 - 580$

$f_1 + f_2 = 106 \quad \dots \text{(Eq. 1)}$

2. Cumulative Frequency (CF) Table

Class Interval Frequency (f) Cumulative Frequency (cf)

$10-20$ $180$ $180$

$20-30$ $f_1$ $180 + f_1$

$30-40$ $34$ $214 + f_1$

$40-50$ $180$ $394 + f_1$

$50-60$ $136$ $530 + f_1$

$60-70$ $f_2$ $530 + f_1 + f_2$

$70-80$ $50$ $580 + f_1 + f_2 = 686$

3. Identifying Median Class

Since the median is $42.6$ , the Median Class is $40 - 50$ .

$l = 40$

$f = 180$

$cf = 214 + f_1$ (CF of the class before the median class)

$h = 10$

$N = 686 \implies \frac{N}{2} = 343$

4. Applying the Median Formula

$\text{Median} = l + \left[ \frac{\frac{N}{2} - cf}{f} \right] \times h$

$42.6 = 40 + \left[ \frac{343 - (214 + f_1)}{180} \right] \times 10$

$2.6 = \frac{129 - f_1}{18}$

$2.6 \times 18 = 129 - f_1$

$46.8 = 129 - f_1$

$f_1 = 129 - 46.8$

$f_1 = 82.2$

Since frequency must be an integer, we round $f_1$ to $82$ .

5. Finding $f_2$

Using Eq. 1:

$f_1 + f_2 = 106$

$82 + f_2 = 106$

$f_2 = 106 - 82$

$f_2 = 24$

Final Answer

The values are $f_1 = 82$ and $f_2 = 24$ .

NIMCET 2021 — Mathematics PYQ

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Class Interval	Frequency (f)	Cumulative Frequency (cf)
$10-20$	$180$	$180$
$20-30$	$f_1$	$180 + f_1$
$30-40$	$34$	$214 + f_1$
$40-50$	$180$	$394 + f_1$
$50-60$	$136$	$530 + f_1$
$60-70$	$f_2$	$530 + f_1 + f_2$
$70-80$	$50$	$580 + f_1 + f_2 = 686$