Median of Grouped Data Questions in English

(B) To calculate the median height,we first convert the cumulative frequency distribution into a standard frequency distribution table.

Class intervals	Frequency $(f)$	Cumulative frequency $(cf)$
Below $140$	$4$	$4$
$140-145$	$7$	$11$
$145-150$	$18$	$29$
$150-155$	$11$	$40$
$155-160$	$6$	$46$
$160-165$	$5$	$51$

Here,$n = 51$. So,$\frac{n}{2} = \frac{51}{2} = 25.5$.
The cumulative frequency just greater than $25.5$ is $29$,which corresponds to the class interval $145-150$. Thus,the median class is $145-150$.
Here,$l = 145$,$cf = 11$,$f = 18$,and $h = 5$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$\text{Median} = 145 + \left( \frac{25.5 - 11}{18} \right) \times 5$
$\text{Median} = 145 + \left( \frac{14.5}{18} \right) \times 5 = 145 + \frac{72.5}{18} = 145 + 4.027... \approx 149.03 \ cm$.

(X=9, Y=15) First,we construct the cumulative frequency table:

Class intervals	Frequency	Cumulative frequency
$0-100$	$2$	$2$
$100-200$	$5$	$7$
$200-300$	$x$	$7+x$
$300-400$	$12$	$19+x$
$400-500$	$17$	$36+x$
$500-600$	$20$	$56+x$
$600-700$	$y$	$56+x+y$
$700-800$	$9$	$65+x+y$
$800-900$	$7$	$72+x+y$
$900-1000$	$4$	$76+x+y$

Given that the total frequency $n = 100$,we have:
$76 + x + y = 100 \implies x + y = 24$ ........... $(1)$
The median is $525$,which lies in the class $500-600$. Thus,$l = 500$,$f = 20$,$cf = 36 + x$,and $h = 100$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100$
$25 = (14 - x) \times 5$
$5 = 14 - x$
$x = 9$
Substituting $x = 9$ into equation $(1)$:
$9 + y = 24 \implies y = 15$.
Thus,$x = 9$ and $y = 15$.

(A) The cumulative frequency for the given data is calculated as follows:

Class interval	Frequency	Cumulative frequency
$0-10$	$5$	$5$
$10-20$	$x$	$5+x$
$20-30$	$20$	$25+x$
$30-40$	$15$	$40+x$
$40-50$	$y$	$40+x+y$
$50-60$	$5$	$45+x+y$

From the table,the total frequency $n = 60$.
Thus,$45+x+y = 60 \implies x+y = 15 \dots (1)$.
The median is $28.5$,which lies in the class interval $20-30$.
Therefore,the median class is $20-30$.
Lower limit $(l) = 20$,frequency $(f) = 20$,cumulative frequency $(cf) = 5+x$,and class size $(h) = 10$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$.
$28.5 = 20 + \left( \frac{30 - (5+x)}{20} \right) \times 10$.
$8.5 = \frac{25-x}{2}$.
$17 = 25 - x \implies x = 8$.
Substituting $x=8$ into equation $(1)$: $8+y = 15 \implies y = 7$.
Thus,$x = 8$ and $y = 7$.

(B) The given data is in the form of 'less than' cumulative frequency distribution. We first convert it into class intervals with their respective frequencies.

Age (in years)	Frequency $(f_i)$	Cumulative frequency $(cf)$
$18-20$	$2$	$2$
$20-25$	$6-2=4$	$6$
$25-30$	$24-6=18$	$24$
$30-35$	$45-24=21$	$45$
$35-40$	$78-45=33$	$78$
$40-45$	$89-78=11$	$89$
$45-50$	$92-89=3$	$92$
$50-55$	$98-92=6$	$98$
$55-60$	$100-98=2$	$100$

Here,$n = 100$,so $\frac{n}{2} = 50$.
The cumulative frequency just greater than $50$ is $78$,which corresponds to the class interval $35-40$.
Thus,the median class is $35-40$.
Lower limit $(l)$ $= 35$,class size $(h)$ $= 5$,frequency of median class $(f)$ $= 33$,and cumulative frequency of the class preceding the median class $(cf)$ $= 45$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$\text{Median} = 35 + \left( \frac{50 - 45}{33} \right) \times 5$
$\text{Median} = 35 + \left( \frac{5}{33} \right) \times 5 = 35 + \frac{25}{33} \approx 35 + 0.76 = 35.76$.
Therefore,the median age is $35.76$ years.

(C) The given data does not have continuous class intervals. The difference between the upper limit of one class and the lower limit of the next class is $127 - 126 = 1$. Therefore,we subtract $0.5$ from the lower limit and add $0.5$ to the upper limit of each class to make them continuous.
Continuous class intervals and cumulative frequencies are as follows:

Length (in mm)	Frequency $(f_i)$	Cumulative Frequency $(cf)$
$117.5-126.5$	$3$	$3$
$126.5-135.5$	$5$	$8$
$135.5-144.5$	$9$	$17$
$144.5-153.5$	$12$	$29$
$153.5-162.5$	$5$	$34$
$162.5-171.5$	$4$	$38$
$171.5-180.5$	$2$	$40$

Here,$n = 40$,so $\frac{n}{2} = 20$. The cumulative frequency just greater than $20$ is $29$,which corresponds to the class interval $144.5-153.5$.
Median class $= 144.5-153.5$,lower limit $(l)$ $= 144.5$,class size $(h)$ $= 9$,frequency $(f)$ $= 12$,and cumulative frequency of the preceding class $(cf)$ $= 17$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$\text{Median} = 144.5 + \left( \frac{20 - 17}{12} \right) \times 9 = 144.5 + \left( \frac{3}{12} \right) \times 9 = 144.5 + 2.25 = 146.75$.
Thus,the median length of the leaves is $146.75 \text{ mm}$.

(D) The cumulative frequencies with their respective class intervals are as follows:

Life time	Number of lamps $(f_i)$	Cumulative frequency $(cf)$
$1500-2000$	$14$	$14$
$2000-2500$	$56$	$70$
$2500-3000$	$60$	$130$
$3000-3500$	$86$	$216$
$3500-4000$	$74$	$290$
$4000-4500$	$62$	$352$
$4500-5000$	$48$	$400$

Here,$n = 400$,so $\frac{n}{2} = 200$.
The cumulative frequency just greater than $200$ is $216$,which corresponds to the class interval $3000-3500$.
Thus,the median class is $3000-3500$.
Lower limit $(l)$ of the median class = $3000$.
Frequency $(f)$ of the median class = $86$.
Cumulative frequency $(cf)$ of the class preceding the median class = $130$.
Class size $(h)$ = $500$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$\text{Median} = 3000 + \left( \frac{200 - 130}{86} \right) \times 500$
$\text{Median} = 3000 + \left( \frac{70}{86} \right) \times 500$
$\text{Median} = 3000 + \frac{35000}{86} \approx 3000 + 406.98 = 3406.98$.
Therefore,the median lifetime of a lamp is $3406.98 \text{ hours}$.

(C) The cumulative frequencies with their respective class intervals are as follows:

Weight (in kg)	Frequency $(f_i)$	Cumulative frequency $(cf)$
$40-45$	$2$	$2$
$45-50$	$3$	$5$
$50-55$	$8$	$13$
$55-60$	$6$	$19$
$60-65$	$6$	$25$
$65-70$	$3$	$28$
$70-75$	$2$	$30$

Total number of students $n = 30$.
We calculate $\frac{n}{2} = \frac{30}{2} = 15$.
The cumulative frequency just greater than $15$ is $19$,which corresponds to the class interval $55-60$.
Thus,the median class is $55-60$.
Lower limit of the median class $(l)$ = $55$.
Frequency of the median class $(f)$ = $6$.
Cumulative frequency of the class preceding the median class $(cf)$ = $13$.
Class size $(h)$ = $5$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$.
$\text{Median} = 55 + \left( \frac{15 - 13}{6} \right) \times 5$.
$\text{Median} = 55 + \left( \frac{2}{6} \right) \times 5 = 55 + \frac{10}{6} = 55 + 1.67 = 56.67$.
Therefore,the median weight of the students is $56.67 \text{ kg}$.

(17.5) To draw the 'more than' ogive,we plot the points $(5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7),$ and $(35, 3)$ on a graph where the horizontal axis represents the lower limits of profit and the vertical axis represents the cumulative frequency. We join these points with a smooth curve.
Next,we convert the given data into a frequency distribution table:

Classes	$5-10$	$10-15$	$15-20$	$20-25$	$25-30$	$30-35$	$35-40$
No. of shops	$2$	$12$	$2$	$4$	$3$	$4$	$3$
Cumulative frequency	$2$	$14$	$16$	$20$	$23$	$27$	$30$

To draw the 'less than' ogive,we plot the points $(10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27),$ and $(40, 30)$ on the same axes.
The abscissa of the point of intersection of the two ogives gives the median. From the graph,the point of intersection is at $x = 17.5$. Therefore,the median profit is $Rs. 17.5$ lakhs.

(N/A) The given cumulative frequency distributions of less than type are:

Weight (in kg) (upper class limits)	Number of students (cumulative frequency)
Less than $38$	$0$
Less than $40$	$3$
Less than $42$	$5$
Less than $44$	$9$
Less than $46$	$14$
Less than $48$	$28$
Less than $50$	$32$
Less than $52$	$35$

Taking upper class limits on the $x$-axis and their respective cumulative frequencies on the $y$-axis,the ogive is drawn as shown in the figure.
Here,$n = 35$.
So,$\frac{n}{2} = 17.5$.
Mark the point $A$ on the curve whose ordinate is $17.5$. The $x$-coordinate corresponding to this point is $46.5$. Therefore,the median of this data is $46.5$.
To verify using the formula,we first determine the class intervals and their frequencies:

Weight (in kg)	Frequency $(f)$	Cumulative frequency $(cf)$
$38-40$	$3-0=3$	$3$
$40-42$	$5-3=2$	$5$
$42-44$	$9-5=4$	$9$
$44-46$	$14-9=5$	$14$
$46-48$	$28-14=14$	$28$
$48-50$	$32-28=4$	$32$
$50-52$	$35-32=3$	$35$

The cumulative frequency just greater than $\frac{n}{2} = 17.5$ is $28$,which belongs to the class interval $46-48$.
Median class $= 46-48$.
Lower class limit $(l) = 46$.
Frequency $(f) = 14$.
Cumulative frequency of the class preceding the median class $(cf) = 14$.
Class size $(h) = 2$.
Using the median formula:
$\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$= 46 + \left( \frac{17.5 - 14}{14} \right) \times 2$
$= 46 + \left( \frac{3.5}{14} \right) \times 2$
$= 46 + \frac{7}{14} = 46 + 0.5 = 46.5$.
The result is verified.

(N/A) The cumulative frequency distribution of 'more than type' can be obtained as follows:

Production yield (lower class limits)	Cumulative frequency
More than or equal to $50$	$100$
More than or equal to $55$	$100 - 2 = 98$
More than or equal to $60$	$98 - 8 = 90$
More than or equal to $65$	$90 - 12 = 78$
More than or equal to $70$	$78 - 24 = 54$
More than or equal to $75$	$54 - 38 = 16$

By plotting the points $(50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16)$ on a graph paper,taking the lower class limits on the $x$-axis and their respective cumulative frequencies on the $y$-axis,we obtain the 'more than type' ogive.

(A) cumulative frequency table is a statistical tool used to organize data by summing the frequencies of each class interval up to a certain point.
This table is specifically required to calculate the median of a grouped frequency distribution.
The formula for the median is $Median = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$,where $cf$ represents the cumulative frequency of the class preceding the median class.
While mean and mode can be calculated using simple frequency tables,the cumulative frequency is essential for the median.

(B) The given data is a 'more than' cumulative frequency distribution.
To find the number of families in the income range $16000-19000$,we subtract the number of families having income more than $19000$ from the number of families having income more than $16000$.
Number of families with income $> 16000 = 69$.
Number of families with income $> 19000 = 50$.
Number of families in the range $16000-19000 = 69 - 50 = 19$.
Therefore,the correct option is $B$.

(C) The cumulative frequency curves are also known as ogives. The 'less than' type ogive and 'more than' type ogive are plotted on the same graph. The point where these two curves intersect represents the median of the grouped data. The $x$-coordinate (abscissa) of this intersection point corresponds to the median value of the data set.

(A) To find the median class,we first convert the discontinuous class intervals into continuous ones by adjusting the boundaries. The difference between the upper limit of one class and the lower limit of the next is $1$. Thus,we subtract $0.5$ from the lower limit and add $0.5$ to the upper limit.

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$-0.5-5.5$	$13$	$13$
$5.5-11.5$	$10$	$23$
$11.5-17.5$	$15$	$38$
$17.5-23.5$	$8$	$46$
$23.5-29.5$	$11$	$57$

The total frequency $N = 57$. The median position is $N/2 = 57/2 = 28.5$. The cumulative frequency just greater than $28.5$ is $38$,which corresponds to the class interval $11.5-17.5$. Therefore,the median class is $11.5-17.5$. The upper limit of this class is $17.5$.

(D) To find the number of athletes who completed the race in less than $14.6\, s$,we need to sum the frequencies of all classes that are less than $14.6$.
The classes less than $14.6$ are:
$13.8-14.0$ (Frequency = $2$)
$14.0-14.2$ (Frequency = $4$)
$14.2-14.4$ (Frequency = $5$)
$14.4-14.6$ (Frequency = $71$)
Total number of athletes $= 2 + 4 + 5 + 71 = 82$.

(B) No,it is not correct to say that an ogive is a graphical representation of a frequency distribution.
An ogive is specifically a graphical representation of a $cumulative$ frequency distribution.
$A$ frequency distribution is typically represented by a histogram,frequency polygon,or frequency curve.
Therefore,an ogive is used to represent the cumulative frequency,not the simple frequency distribution.

(N/A) No,this is not a correct statement. The median of ungrouped data and the median calculated from grouped data are not always the same. This is because the formula used for calculating the median of grouped data,$\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$,is based on the assumption that the observations within each class interval are uniformly distributed. In reality,the actual distribution of data within classes may not be uniform,leading to a slight difference between the two values.

(N/A) To construct the cumulative frequency distribution,we add the frequency of each class to the sum of the frequencies of all preceding classes.

Class	Frequency	Cumulative frequency
$12.5-17.5$	$2$	$2$
$17.5-22.5$	$22$	$2 + 22 = 24$
$22.5-27.5$	$19$	$24 + 19 = 43$
$27.5-32.5$	$14$	$43 + 14 = 57$
$32.5-37.5$	$13$	$57 + 13 = 70$

(B)

Marks (Class)	Frequency $(f)$	Cumulative Frequency $(cf)$
$30-35$	$14$	$14$
$35-40$	$16$	$30$
$40-45$	$18$	$48$
$45-50$	$23$	$71$
$50-55$	$18$	$89$
$55-60$	$8$	$97$
$60-65$	$3$	$100$

Here,$n = 100$.
Therefore,$\frac{n}{2} = \frac{100}{2} = 50$.
The cumulative frequency just greater than $50$ is $71$,which corresponds to the class interval $45-50$. Thus,the median class is $45-50$.
$l$ (lower limit of median class) $= 45$
$cf$ (cumulative frequency of class preceding median class) $= 48$
$f$ (frequency of median class) $= 23$
$h$ (class size) $= 5$
Median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$\text{Median} = 45 + \left( \frac{50 - 48}{23} \right) \times 5$
$\text{Median} = 45 + \left( \frac{2}{23} \right) \times 5$
$\text{Median} = 45 + \frac{10}{23} \approx 45 + 0.4347 \approx 45.43$
Rounding to one decimal place,the median percentage of marks is $45.4$.

(N/A) To construct a cumulative frequency distribution (less than type),we add the frequencies of all classes preceding the upper limit of the current class.

Weight (in $kg$)	Cumulative Frequency
Less than $45$	$4$
Less than $50$	$4 + 4 = 8$
Less than $55$	$8 + 13 = 21$
Less than $60$	$21 + 5 = 26$
Less than $65$	$26 + 6 = 32$
Less than $70$	$32 + 5 = 37$
Less than $75$	$37 + 2 = 39$
Less than $80$	$39 + 1 = 40$

(A) To find the unknown entries,we use the definition of cumulative frequency,where each entry is the sum of the frequency of the current class and all previous classes.
$1$. For the first class $(150-155)$,the cumulative frequency is equal to the frequency: $a = 12$.
$2$. For the second class $(155-160)$,the cumulative frequency is $12 + b = 25$. Thus,$b = 25 - 12 = 13$.
$3$. For the third class $(160-165)$,the cumulative frequency is $c = 12 + b + 10 = 12 + 13 + 10 = 35$.
$4$. For the fourth class $(165-170)$,the cumulative frequency is $c + d = 43$. Substituting $c = 35$,we get $35 + d = 43$,so $d = 43 - 35 = 8$.
$5$. For the fifth class $(170-175)$,the cumulative frequency is $43 + e = 48$. Thus,$e = 48 - 43 = 5$.
$6$. For the sixth class $(175-180)$,the cumulative frequency is $f = 48 + 2 = 50$. Thus,$f = 50$.
Therefore,the values are: $a = 12, b = 13, c = 35, d = 8, e = 5, f = 50$.

(B) First,we construct a cumulative frequency table:

Weekly income (in $Rs.$)	Number of families $(f_i)$	Cumulative frequency $(cf)$
$0-1000$	$250$	$250$
$1000-2000$	$190$	$440$
$2000-3000$	$100$	$540$
$3000-4000$	$40$	$580$
$4000-5000$	$15$	$595$
$5000-6000$	$5$	$600$

Given $n = 600$,we find $\frac{n}{2} = \frac{600}{2} = 300$.
The cumulative frequency just greater than $300$ is $440$,which corresponds to the median class $1000-2000$.
Here,lower limit $l = 1000$,frequency $f = 190$,cumulative frequency of the preceding class $cf = 250$,and class size $h = 1000$.
Using the median formula:
$\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$= 1000 + \left( \frac{300 - 250}{190} \right) \times 1000$
$= 1000 + \left( \frac{50}{190} \right) \times 1000$
$= 1000 + \frac{5000}{19}$
$= 1000 + 263.157... \approx 1263.16$
Rounding to two decimal places,the median income is $Rs. 1263.15$.

(C) Given total number of players $n = 33$.
$\therefore \frac{n}{2} = \frac{33}{2} = 16.5$.
To find the median class,we look at the cumulative frequencies:
- $85-100$: $11$
- $100-115$: $11 + 9 = 20$
Since $16.5$ lies in the cumulative frequency $20$,the median class is $100-115$.
Here,lower limit $(l) = 100$,frequency $(f) = 9$,cumulative frequency of the class preceding the median class $(cf) = 11$,and class width $(h) = 15$.
Using the median formula:
$\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$= 100 + \left( \frac{16.5 - 11}{9} \right) \times 15$
$= 100 + \left( \frac{5.5}{9} \right) \times 15$
$= 100 + \frac{82.5}{9} = 100 + 9.166... \approx 109.17$.
Hence,the median bowling speed is $109.17 \, km/h$.

(D) To find the median,we first prepare the cumulative frequency table:

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$65-85$	$4$	$4$
$85-105$	$5$	$9$
$105-125$	$13$	$22$
$125-145$	$20$	$42$
$145-165$	$14$	$56$
$165-185$	$8$	$64$
$185-205$	$4$	$68$

Here,total frequency $n = 68$.
$\therefore \frac{n}{2} = \frac{68}{2} = 34$.
The cumulative frequency just greater than $34$ is $42$,which corresponds to the class interval $125-145$.
Thus,the median class is $125-145$.
Here,$l = 125$,$cf = 22$,$f = 20$,and class size $h = 20$.
Using the median formula: $M = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$M = 125 + \left( \frac{34 - 22}{20} \right) \times 20$
$M = 125 + 12 = 137$.
Therefore,the median of the given frequency distribution is $137$.

(X=8, Y=7)

Class	Frequency $(f)$	Cumulative frequency $(cf)$
$0-10$	$5$	$5$
$10-20$	$x$	$5+x$
$20-30$	$20$	$25+x$
$30-40$	$15$	$40+x$
$40-50$	$y$	$40+x+y$
$50-60$	$5$	$45+x+y$

Given total frequency $n = 60$.
Therefore,$45 + x + y = 60 \implies x + y = 15$ (Equation $1$).
Since $n = 60$,$\frac{n}{2} = 30$.
The median $28.5$ lies in the class $20-30$. Thus,the median class is $20-30$.
Here,$l = 20$,$cf = 5 + x$,$f = 20$,and class size $h = 10$.
Using the median formula: $Median = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$28.5 = 20 + \left( \frac{30 - (5 + x)}{20} \right) \times 10$
$8.5 = \frac{25 - x}{2}$
$17 = 25 - x \implies x = 8$.
Substituting $x = 8$ in Equation $1$: $8 + y = 15 \implies y = 7$.
The missing frequencies are $x = 8$ and $y = 7$.

(B)

Class	Frequency $(f)$	Cumulative frequency $(cf)$
$5-10$	$5$	$5$
$10-15$	$6$	$11$
$15-20$	$15$	$26$
$20-25$	$10$	$36$
$25-30$	$5$	$41$
$30-35$	$4$	$45$
$35-40$	$2$	$47$
$40-45$	$2$	$49$

Here,the total frequency $n = 49$.
$\therefore \frac{n}{2} = \frac{49}{2} = 24.5$.
The cumulative frequency just greater than $24.5$ is $26$,which corresponds to the class interval $15-20$.
Therefore,the median class is $15-20$.
Here,$l = 15$ (lower limit of median class),$cf = 11$ (cumulative frequency of the class preceding the median class),$f = 15$ (frequency of the median class),and $h = 5$ (class size).
Using the median formula: $M = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$M = 15 + \left( \frac{24.5 - 11}{15} \right) \times 5$
$M = 15 + \left( \frac{13.5}{15} \right) \times 5$
$M = 15 + \frac{13.5}{3} = 15 + 4.5 = 19.5$.
Thus,the median of the given frequency distribution is $19.5$.

(C) To find the median,we first calculate the cumulative frequency $(cf)$:

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$30-35$	$1$	$1$
$35-40$	$2$	$3$
$40-45$	$5$	$8$
$45-50$	$10$	$18$
$50-55$	$17$	$35$
$55-60$	$15$	$50$
$60-65$	$9$	$59$
$65-70$	$5$	$64$
$70-75$	$3$	$67$

Total frequency $N = 67$. So,$N/2 = 67/2 = 33.5$.
Looking at the $cf$ column,the median class is $50-55$ because $33.5$ lies in this interval.
Here,lower limit $l = 50$,class size $h = 5$,frequency of median class $f = 17$,and cumulative frequency of the class preceding the median class $cf = 18$.
Using the median formula: $\text{Median} = l + \left( \frac{N/2 - cf}{f} \right) \times h$
$\text{Median} = 50 + \left( \frac{33.5 - 18}{17} \right) \times 5$
$\text{Median} = 50 + \left( \frac{15.5}{17} \right) \times 5$
$\text{Median} = 50 + \frac{77.5}{17} \approx 50 + 4.5588 \approx 54.56$.

(D) To find the median,we first calculate the cumulative frequency $(cf)$:

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$60-70$	$5$	$5$
$70-80$	$15$	$20$
$80-90$	$20$	$40$
$90-100$	$30$	$70$
$100-110$	$20$	$90$
$110-120$	$8$	$98$

Total frequency $N = 98$. So,$N/2 = 98/2 = 49$.
The cumulative frequency just greater than $49$ is $70$,which corresponds to the class interval $90-100$. Thus,the median class is $90-100$.
Here,$l = 90$,$f = 30$,$cf = 40$,and $h = 10$.
Using the formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$\text{Median} = 90 + \left( \frac{49 - 40}{30} \right) \times 10$
$\text{Median} = 90 + \left( \frac{9}{30} \right) \times 10 = 90 + 3 = 93$.

(A) To find the median,we first calculate the cumulative frequency $(cf)$:
| Class | Frequency $(f)$ | Cumulative Frequency $(cf)$ |
| :--- | :--- | :--- |
| $110-120$ | $6$ | $6$ |
| $120-130$ | $25$ | $31$ |
| $130-140$ | $48$ | $79$ |
| $140-150$ | $72$ | $151$ |
| $150-160$ | $116$ | $267$ |
| $160-170$ | $60$ | $327$ |
| $170-180$ | $38$ | $365$ |
| $180-190$ | $22$ | $387$ |
| $190-200$ | $3$ | $390$ |
Total frequency $N = 390$. So,$N/2 = 390/2 = 195$.
The cumulative frequency just greater than $195$ is $267$,which corresponds to the class interval $150-160$.
Thus,the median class is $150-160$.
Here,lower limit $l = 150$,frequency of median class $f = 116$,cumulative frequency of the class preceding the median class $cf = 151$,and class size $h = 10$.
Using the median formula: $\text{Median} = l + \left( \frac{N/2 - cf}{f} \right) \times h$
$\text{Median} = 150 + \left( \frac{195 - 151}{116} \right) \times 10$
$\text{Median} = 150 + \left( \frac{44}{116} \right) \times 10 = 150 + \frac{440}{116} \approx 150 + 3.79 = 153.79$.

(B) To find the median,we first calculate the cumulative frequency $(cf)$:

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$0-20$	$5$	$5$
$20-40$	$15$	$20$
$40-60$	$20$	$40$
$60-80$	$18$	$58$
$80-100$	$2$	$60$

Total frequency $N = 60$. So,$N/2 = 60/2 = 30$.
The cumulative frequency just greater than $30$ is $40$,which corresponds to the median class $40-60$.
Here,lower limit $l = 40$,class size $h = 20$,frequency of median class $f = 20$,and cumulative frequency of the class preceding the median class $cf = 20$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$\text{Median} = 40 + \left( \frac{30 - 20}{20} \right) \times 20$
$\text{Median} = 40 + \left( \frac{10}{20} \right) \times 20 = 40 + 10 = 50$.

(C) To find the median,we first calculate the cumulative frequency $(cf)$:

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$0-10$	$2$	$2$
$10-20$	$6$	$8$
$20-30$	$8$	$16$
$30-40$	$16$	$32$
$40-50$	$20$	$52$
$50-60$	$18$	$70$
$60-70$	$16$	$86$
$70-80$	$14$	$100$

Total frequency $N = 100$. So,$N/2 = 50$.
The cumulative frequency just greater than $50$ is $52$,which corresponds to the class interval $40-50$.
Thus,the median class is $40-50$.
Here,$l = 40$,$f = 20$,$cf = 32$,and $h = 10$.
Using the median formula: $\text{Median} = l + \left( \frac{N/2 - cf}{f} \right) \times h$
$\text{Median} = 40 + \left( \frac{50 - 32}{20} \right) \times 10$
$\text{Median} = 40 + \left( \frac{18}{20} \right) \times 10 = 40 + 9 = 49$.

(D) To find the median,we first calculate the cumulative frequency $(cf)$:

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$0-5$	$5$	$5$
$5-10$	$15$	$20$
$10-15$	$15$	$35$
$15-20$	$8$	$43$
$20-25$	$4$	$47$
$25-30$	$2$	$49$
$30-35$	$1$	$50$

Total frequency $N = 50$. So,$N/2 = 25$.
The cumulative frequency just greater than $25$ is $35$,which corresponds to the class interval $10-15$.
Thus,the median class is $10-15$.
Here,$l = 10$,$f = 15$,$cf$ (of preceding class) $= 20$,and $h = 5$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$\text{Median} = 10 + \left( \frac{25 - 20}{15} \right) \times 5$
$\text{Median} = 10 + \left( \frac{5}{15} \right) \times 5 = 10 + \frac{25}{15} = 10 + 1.666... = 11.67$.

(A) Given,total frequency $N = 100$.
Sum of frequencies: $2 + 5 + x + 12 + 17 + 20 + y + 9 + 7 + 4 = 100$
$76 + x + y = 100 \implies x + y = 24$ --- $(1)$
Since the median is $525$,the median class is $500-600$.
Here,$l = 500$,$f = 20$,$cf = (2 + 5 + x + 12 + 17) = 36 + x$,$h = 100$,and $N/2 = 50$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100$
$25 = (14 - x) \times 5$
$5 = 14 - x \implies x = 9$
Substituting $x = 9$ in $(1)$: $9 + y = 24 \implies y = 15$.
Thus,$x = 9$ and $y = 15$.

(N/A) Given total frequency $N = 230$.
Sum of frequencies: $12 + 30 + x + 65 + y + 25 + 18 = 230$
$150 + x + y = 230 \implies x + y = 80$ ---(Equation $1$)
Since the median is $46$,the median class is $40-50$.
Here,$l = 40$,$h = 10$,$f = 65$,$N = 230$,and cumulative frequency of the class preceding the median class $cf = 12 + 30 + x = 42 + x$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$46 = 40 + \left( \frac{115 - (42 + x)}{65} \right) \times 10$
$6 = \left( \frac{73 - x}{65} \right) \times 10$
$6 \times 6.5 = 73 - x$
$39 = 73 - x \implies x = 34$.
Substituting $x = 34$ in Equation $1$: $34 + y = 80 \implies y = 46$.
Thus,$x = 34$ and $y = 46$.

(A) Given total frequency $N = 60$.
Sum of frequencies: $5 + x + 20 + y + 2 = 60 \implies x + y + 27 = 60 \implies x + y = 33$ (Equation $1$).
Since the median is $50$,the median class is $40-60$. Here,$l = 40$,$h = 20$,$f = 20$,$cf = 5 + x$,and $N/2 = 30$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$.
$50 = 40 + \left( \frac{30 - (5 + x)}{20} \right) \times 20$.
$50 = 40 + (25 - x)$.
$50 = 65 - x \implies x = 15$.
Substituting $x = 15$ in Equation $1$: $15 + y = 33 \implies y = 18$.
Thus,$x = 15$ and $y = 18$.

(A) Given total frequency $N = 100$,so $2 + 6 + 8 + x + 20 + 18 + y + 14 = 100$.
$68 + x + y = 100 \implies x + y = 32$ (Equation $1$).
Since the median is $49$,the median class is $40-50$. Here $l = 40$,$f = 20$,$cf = 2 + 6 + 8 + x = 16 + x$,and $h = 10$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$.
$49 = 40 + \left( \frac{50 - (16 + x)}{20} \right) \times 10$.
$9 = \frac{34 - x}{2} \implies 18 = 34 - x \implies x = 16$.
Substituting $x = 16$ in Equation $1$: $16 + y = 32 \implies y = 16$.
Thus,$x = 16$ and $y = 16$.

(B) In the formula for the median of grouped data,$M = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$,the term $l$ represents the lower limit of the median class.

(C) In the formula for the median,$M = l + \frac{(\frac{n}{2} - cf)}{f} \times h$,the term $n$ represents the total frequency of the given data distribution,which is the sum of all individual frequencies,i.e.,$n = \sum f_i$.

(A) In the median formula,$cf$ represents the cumulative frequency of the class preceding the median class. Here,$l$ is the lower limit of the median class,$n$ is the total number of observations,$f$ is the frequency of the median class,and $c$ is the class size (or class length).

(B) In the formula for the median,$M = l + \frac{(\frac{n}{2} - cf)}{f} \times h$,the term $f$ represents the frequency of the median class.

(C) In the formula for the median of grouped data,$M = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$,the term $h$ represents the class size or class length of the median class. Therefore,$h = \text{class length}$.

(D) To find the median class,we first calculate the cumulative frequency $(cf)$ for the given data:
$1$. Class $0-10$: Frequency $12$,$cf = 12$
$2$. Class $10-20$: Frequency $18$,$cf = 12 + 18 = 30$
$3$. Class $20-30$: Frequency $20$,$cf = 30 + 20 = 50$
$4$. Class $30-40$: Frequency $17$,$cf = 50 + 17 = 67$
$5$. Class $40-50$: Frequency $13$,$cf = 67 + 13 = 80$
The total number of observations $N = 80$.
We find $N/2 = 80/2 = 40$.
The median class is the class interval whose cumulative frequency is greater than or equal to $N/2 = 40$.
Looking at the cumulative frequencies $(12, 30, 50, 67, 80)$,the first value greater than $40$ is $50$.
Therefore,the corresponding class interval is $20-30$.

(C) cumulative frequency curve is a graphical representation of the cumulative frequency distribution.
It is plotted by taking the upper class limits on the $x$-axis and the corresponding cumulative frequencies on the $y$-axis.
This curve is commonly referred to as an $ogive$.

(B) The median of grouped data is calculated using the formula:
$M = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
Where:
$l$ = lower limit of the median class
$n$ = total number of observations
$cf$ = cumulative frequency of the class preceding the median class
$f$ = frequency of the median class
$h$ = class size (width of the class interval).

(C) The cumulative frequency of a class is the sum of the frequencies of all classes up to that class.
Cumulative frequency of the third class = (Frequency of first class) + (Frequency of second class) + (Frequency of third class)
Cumulative frequency of the third class = $8 + 15 + 18 = 41$.

(D) The cumulative frequency of a class is the sum of the frequencies of all classes up to that class.
Let $cf_n$ be the cumulative frequency of the $n^{th}$ class and $f_n$ be the frequency of the $n^{th}$ class.
We know that $cf_n = cf_{n-1} + f_n$.
Therefore,$cf_{n-1} = cf_n - f_n$.
Given that the cumulative frequency of the fourth class $(cf_4)$ is $25$ and the frequency of the fourth class $(f_4)$ is $10$,we have:
$cf_3 = cf_4 - f_4$
$cf_3 = 25 - 10 = 15$.
Thus,the cumulative frequency of the third class is $15$.

(A) First,calculate the cumulative frequency $(cf)$ for each class:
- $20-25$: $cf = 2$
- $25-30$: $cf = 2 + 5 = 7$
- $30-35$: $cf = 7 + 8 = 15$
- $35-40$: $cf = 15 + 10 = 25$
- $40-45$: $cf = 25 + 7 = 32$
- $45-50$: $cf = 32 + 10 = 42$
- $50-55$: $cf = 42 + 3 = 45$
The total frequency $n = 45$.
We need to find $\frac{n}{2} = \frac{45}{2} = 22.5$.
The median class is the class whose cumulative frequency is just greater than $22.5$.
Looking at the cumulative frequencies,$25$ is the first value greater than $22.5$,which corresponds to the class interval $35-40$.
Therefore,the median class is $35-40$.

(D) First,calculate the cumulative frequency $(cf)$ for each class:
$1$. Class $10-25$: $cf = 5$
$2$. Class $25-40$: $cf = 5 + 21 = 26$
$3$. Class $40-55$: $cf = 26 + 21 = 47$
$4$. Class $55-70$: $cf = 47 + 8 = 55$
$5$. Class $70-85$: $cf = 55 + 25 = 80$
$6$. Class $85-100$: $cf = 80 + 20 = 100$
The total frequency $n = 100$.
Calculate $\frac{n}{2} = \frac{100}{2} = 50$.
The median class is the class whose cumulative frequency is just greater than or equal to $\frac{n}{2} = 50$.
Looking at the cumulative frequencies: $5, 26, 47, 55, 80, 100$. The first cumulative frequency greater than $50$ is $55$,which corresponds to the class $55-70$.
Therefore,the median class is $55-70$.

(A) To find the median class,we first determine the total frequency $n = \sum f_i = 6 + 10 + 5 + 6 + 4 + 2 + 2 = 35$.
The median class is the class interval corresponding to the cumulative frequency just greater than $\frac{n}{2}$.
Here,$\frac{n}{2} = \frac{35}{2} = 17.5$.
Let's calculate the cumulative frequencies $(cf)$:
Class $10-20$: $cf = 6$
Class $20-30$: $cf = 6 + 10 = 16$
Class $30-40$: $cf = 16 + 5 = 21$
Since $21$ is the first cumulative frequency greater than $17.5$,the corresponding class interval $30-40$ is the median class.

(B) To find the median class,we first calculate the cumulative frequency $(cf)$ for each class:
$1$. For class $0-10$,$cf = 4$.
$2$. For class $10-20$,$cf = 4 + 5 = 9$.
$3$. For class $20-30$,$cf = 9 + 9 = 18$.
$4$. For class $30-40$,$cf = 18 + 10 = 28$.
$5$. For class $40-50$,$cf = 28 + 14 = 42$.
$6$. For class $50-60$,$cf = 42 + 8 = 50$.
The total number of observations is $n = 50$.
We need to find the class where the cumulative frequency is greater than or equal to $\frac{n}{2} = \frac{50}{2} = 25$.
Looking at the cumulative frequencies: $4, 9, 18, 28, 42, 50$.
The first cumulative frequency greater than $25$ is $28$,which corresponds to the class interval $30-40$.
Therefore,the median class is $30-40$.