Class 10 Statistics - Median of Grouped Data

Q: The median of the following data is \$525\$. Find the values of \$x\$ and \$y\$, if the total frequency is \$100 \$Class intervalFrequency\$0-100\$\$2\$\$100-200\$\$5\$\$200-300\$\$x\$\$300-400\$\$12\$\$400-500\$\$17\$\$500-600\$\$20\$\$600-700\$\$y\$\$700-800\$\$9\$\$800-900\$\$7\$\$900-1000\$\$4\$

Computation of MedianClass intervalFrequency (f)Cumulative frequency (cf)0-10022100-20057200-300x7+x300-4001219+x400-5001736+x500-6002056+x600-700y56+x+y700-800965+x+y800-900772+x + y900-1000476+x + yTotal = 100We have,\$N= \sum f_i= 100 \$\$\Rightarrow 76 + x + y = 100 \Rightarrow x+ y = 24\$It is given that the median is \$525\$. Clearly, it lies in the class \$500-600\$ \$\therefore l= 500, h= 100, f= 20, F= 36+ x \$ and \$ N= 100\$Now,\$Median = i+ \frac{\frac{N}{2}-F}{f}\times h\$\$\Rightarrow 525 = 500 +\frac{50-(36+x)}{20}\times 100\$\$\Rightarrow 525 -500= (14- x)\times 5\$\$\Rightarrow 25= 70-5x \Rightarrow 5x = 45 \Rightarrow x=9\$Putting \$x=9 \$ in\$ x+y= 24\$, we get \$y= 15.\$Hence, \$x= 9 \$and \$y= 15.\$

Q: Given the following distribution table:Class\$0-5\$\$5-10\$\$10-15\$\$15-20\$\$20-25\$Frequency\$10\$\$15\$\$12\$\$20\$\$9\$The sum of lower limits of the Median class and the Modal class is:

Answer:-The cumulative frequency Table is:- ClassFrequency Cumulative frequency 0-5 10 10 5-10 15 25 10-15 12 37 15-20 20 57 20-25 9 66Here, \$ \cfrac{N}{2} = \cfrac{66}{2} = 33 \$Thus \$37\$ is just greater than \$33\$, \$\therefore\$ The median class is "\$10-15.\$"The lower limit of the median class is 10.Modal class = class with maximum frequency.\$ \therefore \$ Modal class is \$15-20.\$The lower limit of the modal class is \$15\$.Sum of lower limits of median class and modal class = \$10 + 15 = 25\$

Q: If the difference of mode and median of a data is 24 then the difference of median and mean is

Since, Mode = 3 Median - 2 Mean \$\Rightarrow \$ Mode = 2 Median + Median - 2 Mean \$\Rightarrow \$ Mode - Median = 2 Median - 2 Mean \$\Rightarrow \$ Mode - Median = 2 (Median - Mean)Given, Difference of mode and median of a data is 24\$\therefore \$ 24 = 2 (Median - Mean)\$\Rightarrow \$ Median - Mean = 12\$\therefore \$ Option A is correct.

Q: Which of the following is not a measure of central tendency?

\${\textbf{Step -1: Defining all terms}}\$ \$\text{Mean is the average }\$ \${\text{Median is the middle number,when data is arranged in numerical order}}{\text{.}}\$ \${\text{Mode is the data item that appears most frequently}}{\text{.}}\$ \${\textbf{Hence, mean, median and mode are all measures of central tendencies}}{\text{.}}\$\$\textbf{Therefore, correct answer is option D.}\$

Q: Calculate the missing frequency from the following distribution, it being given that the median of the distribution is \$24\$.Age in years0-1010-2020-3030-4040-50No. of persons\$5\$\$25\$?\$18\$\$7\$

Let the missing frequency be \$x\$.Age in years(x) No. of persons(f) C.F.0-105510-20253020-30\$x\$30 \$+ x\$30-401848 \$+ x\$40-50755 \$+ x\$ \$N = 55 + x\$As, \$N = 55 + x\$ \$\Rightarrow \cfrac{N}{2} = \cfrac{55 + x}{2}\$\$Median = l + \cfrac{\cfrac{N}{2} + C}{f} \times h\$As \$median\$ is given \$24\$, therefore median class is \$20-30.\$Therefore,\$l = \$lower limit of median class \$= 20\$\$C = C.F. \$ of the class preceding the median class \$= 30\$\$h =\$ higher limit - lower limit \$= 40 - 30 = 10\$\$f = \$frequency of median class \$= x\$\$\therefore \; median = 20 + \cfrac{\cfrac{55 + x}{2} - 30}{x} \times 10\$\$24 = 20 + \cfrac{x - 5}{2x} \times 10\$\$4 = \cfrac{x - 5}{x} \times 5\$\$\Rightarrow 4x = 5x - 25\$\$\Rightarrow x = 25\$Hence, the missing frequency is \$25.\$

Q: A life insurance agent found the following data for the distribution of ages of \$100\$ policy holders. Calculate the median age, if policies are only given to persons having age \$18\$ years on wards but less than \$60\$ years.Age in yearsNumber of policy holdersBelow 20\$2\$Below 25\$6\$Below 30\$24\$Below 35\$45\$Below 40\$78\$Below 45\$89\$Below 50\$92\$Below 55\$98\$Below 60\$100\$

Class intervalFrequencyCumulative frequency \$15-20\$\$2\$ \$2\$ \$20-25\$\$4\$ \$6\$ \$25-30\$\$18\$ \$24\$ \$30-35\$\$21\$ \$45\$ \$35-40\$\$33\$ \$78\$ \$40-45\$\$11\$ \$89\$ \$45-50\$\$3\$ \$92\$ \$50-55\$\$6\$ \$98\$ \$55-60\$\$2\$ \$100\$ \$\Rightarrow\$ \$N=100\$ then, \$\dfrac{N}{2}=50\$The cumulative frequency just greater then \$50\$ is \$78\$.\$\therefore\$ Median class \$=35-40\$Here, \$l=35,\,h=5,\,f=33,\,cf=45\$\$Median=l+\dfrac{\dfrac{N}{2}-cf}{f}\times h\$\$\Rightarrow\$ \$Median=35+\dfrac{50-45}{33}\times 5\$\$\Rightarrow\$ \$Median=35+\dfrac{25}{33}=35.76\$

Q: Find the median class of the following distribution:Class\$0-10\$\$10-20\$\$20-30\$\$30-40\$\$40-50\$\$50-60\$\$60-70\$Frequency\$4\$\$4\$\$8\$\$10\$\$12\$\$8\$\$4\$

First we find the cumulative frequencyClassesFrequencyCumulative Frequency\$0-10\$\$4\$\$4\$\$10-20\$\$4\$\$8\$\$20-30\$\$8\$\$16\$\$30-40\$\$10\$\$26\$\$40-50\$\$12\$\$38\$\$50-60\$\$8\$\$46\$\$60-70\$\$4\$\$50\$Total\$50\$Here, \$\dfrac{n}{2}=\dfrac{50}{2}=25\$\$\therefore\$ Median class \$=30-40\$.

Q: Find the median class for the following data given below\$:\$Number of cars\$0-10\$\$10-20\$\$20-30\$\$30-40\$\$40-50\$\$50-60\$\$60-70\$\$70-80\$Frequency\$7\$\$9\$\$13\$\$21\$\$12\$\$15\$\$4\$\$12\$

To find the median class\$:\$First add the total number of frequencies.Here, \$f = 93\$\$= \dfrac{(Total\quad number\quad of \quad frequency)}{2}\$\$= (93)/2 = 46.5\$Hence, the value 46.5 lies in the class interval 30-40 from the cummulative frequency table.

Q: Compute the median for each of the following data:(i)MarksNo. of students (ii)MarksNo. of studentsLess than \$10\$ \$0\$More than \$150\$ \$0\$Less than \$30\$ \$10\$More than \$140\$ \$12\$Less than \$50\$ \$25\$More than \$130\$ \$27\$Less than \$70\$ \$43\$More than \$120\$ \$60\$Less than \$90\$ \$65\$More than \$110\$ \$105\$Less than \$110\$ \$87\$More than \$100\$ \$124\$Less than \$130\$ \$96\$More than \$90\$ \$141\$Less than \$150\$ \$100\$More than \$80\$ \$150\$

\$( i )\$ Class intervalFrequency Cumulative frequency \$0-10\$\$0\$ \$0\$ \$10-30\$\$10\$ \$10\$ \$30-50\$\$15\$ \$25\$ \$50-70\$\$18\$ \$43\$ \$70-90\$\$22\$ \$65\$ \$90-110\$\$22\$ \$87\$ \$110-130\$\$9\$ \$96\$ \$130-150\$\$4\$ \$100\$ \$N=100\$ We have, \$N=100,\,\dfrac{N}{2}=50\$The cumulative frequency just greater than \$\dfrac{N}{2}\$ is \$65\$ then median class is \$70-90\$ such that,\$l=70,\,h=90-70=20,\,f=22,\,cf=43\$\$Median=l+\dfrac{\dfrac{N}{2}-cf}{f}\times h\$\$\Rightarrow\$ \$Median=70+\dfrac{50-43}{22}\times 20\$\$\Rightarrow\$ \$Median=70+\dfrac{7\times 20}{22}\$\$\Rightarrow\$ \$Median=70+6.36\$\$\therefore\$ \$Median=76.36\$\$(ii)\$ Class intervalFrequency Cumulative Frequency \$80-90\$\$9\$ \$9\$ \$90-100\$\$17\$ \$26\$ \$100-110\$\$19\$ \$45\$ \$110-120\$\$45\$ \$90\$ \$120-130\$\$33\$ \$123\$ \$130-140\$\$15\$ \$138\$ \$140-150\$\$12\$ \$150\$ \$150-160\$\$0\$ \$150\$ \$N=150\$ We have, \$N=150,\$ then, \$\dfrac{N}{2}=\dfrac{150}{2}=75\$The cumulative frequency just more than \$\dfrac{N}{2}\$ is \$90\$ then the median class is \$110-120\$ such that \$l=70,\,h=120-110=10,\,f=45,\,cf=45\$\$Median=l+\dfrac{\dfrac{N}{2}-cf}{f}\times h\$\$\Rightarrow\$ \$Median=110+\dfrac{75-45}{45}\times 10\$\$\Rightarrow\$ \$Median=110+\dfrac{30\times 10}{45}\$\$\Rightarrow\$ \$Median=110+6.67\$\$\therefore\$ \$Median=116.67\$

Q: If the median of the following frequency distribution is 46, find the missing frequenciesVariable10 - 2020 - 3030 - 4040 - 5050 - 6060 - 7070 - 80TotalFrequency1230P65Q2518230

Class intervalFrequencyC.F.10 - 20121220 - 30304230 - 40\$\displaystyle f_{1}\$\$\displaystyle 42+f_{1}\$40 - 5065\$\displaystyle 107+f_{1}\$50 - 60\$\displaystyle f_{2}\$\$\displaystyle 107+f_{1}+f_{2}\$60 - 7025\$\displaystyle 132+f_{1}+f_{2}\$70 - 8018\$\displaystyle 150+f_{1}+f_{2}\$Let the frequency of the class 30 - 40 be \$\displaystyle f_{1}\$ and that of the calss 50 - 60 be \$\displaystyle f_{2}\$ The total frequency is 22912 + 30 + \$\displaystyle f_{1}\$ + 65 + \$\displaystyle f_{2}\$ + 25 + 18 = 229 \$\displaystyle \Rightarrow f_{1}+f_{2}=79\$It is given that median is 46 clearly 46 lies in the class 40 - 50. So 40 - 50 is the median class\$\displaystyle \therefore l=40,h=10,f=65\: and\: C=42+f_{1},N=229\$Median = \$\displaystyle l+\dfrac{\dfrac{N}{2}-C}{f}\times h\Rightarrow 46=40+\dfrac{\dfrac{229}{2}-(42+f_{1})}{65}\times 10\Rightarrow 46=40+\frac{145-2f_{1}}{13}\$\$\displaystyle \Rightarrow 6=\dfrac{145-2f_{1}}{13}\rightarrow 2f_{1}=67\Rightarrow f_{1}=33.5\: or\: 34\$ Since \$\displaystyle f_{1}+f_{2}=79\: \: \therefore f_{2}=45\$ Hence \$\displaystyle f_{1}=34\$ and \$\displaystyle f_{2}=45\$

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>> $Median of Grouped Data$

Median of Grouped Data

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Median of Data using Cumulative Frequency Distribution

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Median of Ungrouped and Grouped Data

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Relation between Mean, Median and Mode

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Median of Grouped Data

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Median of discrete grouped data

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If the median of the following frequency distribution is 46, find the missing frequencies

Variable 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 Total
Frequency 12 30 P 65 Q 25 18 230

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Class interval	Frequency
$0 - 100$	$2$
$100 - 200$	$5$
$200 - 300$	$x$
$300 - 400$	$12$
$400 - 500$	$17$
$500 - 600$	$20$
$600 - 700$	$y$
$700 - 800$	$9$
$800 - 900$	$7$
$900 - 1000$	$4$

Class	$0 - 5$	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$
Frequency	$10$	$15$	$12$	$20$	$9$

Age in years	0-10	10-20	20-30	30-40	40-50
No. of persons	$5$	$25$	?	$18$	$7$

Age in years	Number of policy holders
Below 20	$2$
Below 25	$6$
Below 30	$24$
Below 35	$45$
Below 40	$78$
Below 45	$89$
Below 50	$92$
Below 55	$98$
Below 60	$100$

Class	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$
Frequency	$4$	$4$	$8$	$10$	$12$	$8$	$4$

Number of cars	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	$70 - 80$
Frequency	$7$	$9$	$13$	$21$	$12$	$15$	$4$	$12$

(i)Marks	No. of students	(ii)Marks	No. of students
Less than $10$	$0$	More than $150$	$0$
Less than $30$	$10$	More than $140$	$12$
Less than $50$	$25$	More than $130$	$27$
Less than $70$	$43$	More than $120$	$60$
Less than $90$	$65$	More than $110$	$105$
Less than $110$	$87$	More than $100$	$124$
Less than $130$	$96$	More than $90$	$141$
Less than $150$	$100$	More than $80$	$150$

Variable	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80	Total
Frequency	12	30	P	65	Q	25	18	230

Revise with Concepts

Median of Data using Cumulative Frequency Distribution

Quick Summary With Stories

Relation between Mean, Median and Mode

Median of Grouped Data

Median of discrete grouped data

Median of Continuously grouped data

The median of the following data is 525. Find the values of x and y, if the total frequency is 100Class intervalFrequency0−1002100−2005200−300x300−40012400−50017500−60020600−700y700−8009800−9007900−10004

Given the following distribution table: Class 0−5 5−10 10−15 15−20 20−25 Frequency 10 15 12 20 9 The sum of lower limits of the Median class and the Modal class is:

If the difference of mode and median of a data is 24 then the difference of median and mean is

Which of the following is not a measure of central tendency?

Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24. Age in years0-1010-2020-3030-4040-50No. of persons525?187

Find the median class of the following distribution: Class0−1010−2020−3030−4040−5050−6060−70Frequency448101284

Find the median class for the following data given below: Number of cars 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Frequency 7 9 13 21 12 15 4 12

If the median of the following frequency distribution is 46, find the missing frequencies Variable10 - 2020 - 3030 - 4040 - 5050 - 6060 - 7070 - 80TotalFrequency1230P65Q2518230

The median of the following data is $525$ . Find the values of $x$ and $y$ , if the total frequency is $100$
Class interval Frequency
$0 - 100$ $2$
$100 - 200$ $5$
$200 - 300$ $x$
$300 - 400$ $12$
$400 - 500$ $17$
$500 - 600$ $20$
$600 - 700$ $y$
$700 - 800$ $9$
$800 - 900$ $7$
$900 - 1000$ $4$

Given the following distribution table:

Class
$0 - 5$
$5 - 10$
$10 - 15$
$15 - 20$
$20 - 25$

Frequency
$10$
$15$
$12$
$20$
$9$

The sum of lower limits of the Median class and the Modal class is:

Calculate the missing frequency from the following distribution, it being given that the median of the distribution is $24$ .

Age in years 0-10 10-20 20-30 30-40 40-50
No. of persons $5$ $25$ ? $18$ $7$

Find the median class of the following distribution:

Class $0 - 10$ $10 - 20$ $20 - 30$ $30 - 40$ $40 - 50$ $50 - 60$ $60 - 70$
Frequency $4$ $4$ $8$ $10$ $12$ $8$ $4$

Find the median class for the following data given below $:$

Number of cars
$0 - 10$
$10 - 20$
$20 - 30$
$30 - 40$
$40 - 50$
$50 - 60$
$60 - 70$
$70 - 80$

Frequency
$7$
$9$
$13$
$21$
$12$
$15$
$4$
$12$

If the median of the following frequency distribution is 46, find the missing frequencies

Variable 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 Total
Frequency 12 30 P 65 Q 25 18 230