Important Questions for CBSE Class 10 Maths Chapter 14 - Statistics 2024-25

1. Find the median from the following grades of $70$ students:

Ans:

Here,$N=70$

Therefore, $\dfrac{\text{N}}{\text{2}}=\dfrac{70}{2}$

$\dfrac{\text{N}}{\text{2}}=35$

Median can be calculated as,

$Median=\dfrac{1}{2}\left( {{\dfrac{N}{2}}^{th}}term+{{\left( \dfrac{N}{2}+1 \right)}^{th}}term \right)$

Therefore, \[Median=\dfrac{1}{2}\left( 35+\left( 35+1 \right) \right)\]

\[Median=\dfrac{1}{2}\left( {{35}^{th}}term+{{36}^{th}}term \right)\]

\[Median=\dfrac{1}{2}\left( 50+50 \right)\]

\[Median=\dfrac{100}{2}\]

Therefore, 

\[Median=50\]

Therefore, The corresponding value of marks for $35$ is $50$.

2. The sum of deviations for a set of values \[{{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{2}}}\text{,}{{\text{x}}_{\text{3}}}\text{,}...........{{\text{x}}_{\text{n}}}\] measured from $50$ is $-10$, while the sum of deviations for the values from $46$ is $70$. Find the value of \[\text{n}\]and the mean.

Ans:From the given data, 

$\sum\limits_{\text{i = 1}}^{\text{n}}{\text{(}{{\text{X}}_{\text{i}}}}\text{- 50) = 10}$ and 

$\sum\limits_{\text{i = 1}}^{\text{n}}{\text{(}{{\text{X}}_{\text{i}}}}\text{- 46) = 70}$

Therefore,

$\sum\limits_{\text{i = 1}}^{\text{n}}{{{\text{X}}_{\text{i}}}}\text{- 50n = -10    }............\text{(1)}$

And

\[\sum\limits_{\text{i = 1}}^{\text{n}}{{{\text{X}}_{\text{i}}}}\text{- 46m = 70    }............\text{(2)}\]

By subtracting \[\text{(2)}\]from \[\text{(1)}\], we get

\[\text{-4n = -80}\] 

Therefore,

\[\text{n = 20}\]

By putting value of \[\text{n}\] in eq \[\text{(1)}\], we get

 $\sum\limits_{\text{i=1}}^{\text{n}}{{{\text{X}}_{\text{i}}}}\text{-50  }\!\!\times\!\!\text{  20 = -10}$

$\sum\limits_{\text{i=1}}^{\text{n}}{{{\text{X}}_{\text{i}}}}\text{ - 100 = -10}$

Therefore,

$\sum\limits_{\text{i=1}}^{\text{n}}{{{\text{X}}_{\text{i}}}}\text{ = 990}$

Mean can be calculated as,

$\text{Mean = }\dfrac{\text{1}}{\text{n}}\left( \sum\limits_{\text{i=1}}^{\text{n}}{{{\text{X}}_{\text{i}}}} \right)$

Therefore,

$\text{Mean = }\dfrac{\text{990}}{\text{20}}$

$\text{Mean = 49}\text{.5}$

Therefore, 

\[\text{n = 20}\] and 

$\text{Mean = 49}\text{.5}$.

3. Prove: $\sum{({{x}_{8}}}-\overline{x})=0$

Ans: To prove that the algebraic sum of deviation from mean is zero that is

$\sum\limits_{i=1}^{n}{({{X}_{i}}}-\overline{X})=0$

We have,

$\overline{X}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{X}_{i}}}$

$n\overline{X}=\sum\limits_{i=1}^{n}{{{X}_{i}}}$

Now, 

$\sum\limits_{i=1}^{n}{({{X}_{i}}}-\overline{X})=({{X}_{1}}+{{X}_{2}}+.......+{{X}_{n}})-n\overline{X}$

$\sum\limits_{i=1}^{n}{({{X}_{i}}}-\overline{X})=({{X}_{1}}+{{X}_{2}}+.......+{{X}_{n}})-n\overline{X}$

$\sum\limits_{i=1}^{n}{({{X}_{i}}}-\overline{X})=\sum\limits_{i=1}^{n}{{{X}_{i}}}-n\overline{X}$

$\sum\limits_{i=1}^{n}{({{X}_{i}}}-\overline{X})=n\overline{X}-n\overline{X}$

Therefore,

$\sum\limits_{i=1}^{n}{({{X}_{i}}}-\overline{X})=0$

Thus, $\sum\limits_{i=1}^{n}{({{X}_{i}}}-\overline{X})=0$ is proved.

4. Determine the median from the following data:

Ans: Here, the mid values are given and hence, we need to find the upper and lower limits of the various classes. 

The difference between two consecutive values is,

\[\text{h = 125 --115 = 10}\]

Therefore,

Lower limit of a class \[\text{= Mid value - }\dfrac{\text{h}}{\text{2}}\]

Upper limit \[\text{= Mid value + }\dfrac{\text{h}}{\text{2}}\]

From above data, we get

$N=390$

Therefore,

$\dfrac{N}{2}=\dfrac{390}{2}$

$\dfrac{N}{2}=195$ that lies in the class $150-160$

Therefore, Median Class\[=150-160\]

Size of interval, \[h=160-150=10\]

Lower limit of Modal class \[l=150\]

\[CF=151\]

\[\text{Median}=l+\dfrac{\dfrac{N}{2}-CF}{f}\times h\]

Therefore,

$Median=150+\dfrac{195-151}{116}\times 10$

$Median=150+3.79$

$Median=153.79$

5. The mean of ‘$n$’ observation is $\overline{x}$, Find the new mean if the first term is increased by $1$, second by $2$ and so on. 

Ans: Mean can be calculated as;

\[\text{Mean}=\sum\limits_{i=1}^{n}{{{x}_{i}}}=nx\]

Where,

\[{{x}_{i}}=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{n}}}{n}\]

Here, the first term is increased by $1$, second by $2$ and so on that is by adding $1$ to \[{{x}_{1}}\], \[2\] to \[{{x}_{2}}\], and so on, we get

\[{{x}_{1}}+1+{{x}_{2}}+2+{{x}_{3}}+3+...+{{x}_{n}}+n\]

\[\Rightarrow \left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...+{{x}_{n}} \right)+\left( 1+2+3+...+n \right)\]

\[\Rightarrow \sum{n}+\dfrac{n\left( n+1 \right)}{2}\]

The Mean of the new number is 

\[\Rightarrow \dfrac{\sum{n}+\dfrac{n\left( n+1 \right)}{2}}{n}\]

$\Rightarrow \dfrac{2nx+n\left( n+1 \right)}{2n}$

$\Rightarrow x+\dfrac{(n+1)}{2}$

6. Find the median for a frequency distribution, where mode is $7.88$ and mean is $8.32$.

Ans:Mode can be calculated as;

\[\text{Mode = 3 median -- 2 mean}\]

By putting all values in above, we get

\[\text{7}\text{.88 = 3 median -- 2 }\left( 8.32 \right)\]

Therefore,

\[\text{3 Median = 7}\text{.88 + 16}\text{.64 }\]

\[Median=\dfrac{24.52}{3}\]

\[\therefore Median\text{ }=\text{ }8.17\]

Therefore, median is \[8.17\].

7. The mode of a distribution is \[55\] & the modal class is \[45-60\] and the frequency preceding the modal class is \[5\] and the frequency after the modal class is \[10\] . Determine the frequency of the modal class.

Ans:We have,

Mode $=55$

Modal Class $=45-60$

Interval size, $h=60-45=15$

Modal class preceding, ${{f}_{1}}=5$

After the modal class, ${{f}_{2}}=10$

Mode can be calculated as;

$Mode=L+\dfrac{f-{{f}_{1}}}{2f-f1-f2}\times h$

Therefore, by putting all values in above equation, we get

$55=45+\dfrac{f-5}{2f-5-10}\times 15$

$10=\left( \dfrac{f-5}{2f-15} \right)\times 15$

$\dfrac{10}{15}=\dfrac{f-5}{2f-15}$

$20f-150=15f-75$

$5f=75$

$f=\dfrac{75}{5}$

Therefore,

$f=15$

Therefore, the frequency of the modal class is $15$.

8. The mean of $30$ numbers is $18$ . Find the new mean, if each observation is increased by $2$ ?

Ans: The mean of  $30$ numbers is $18$.

Therefore,$\text{Sum of 30 numbers = 30 }\!\!\times\!\!\text{ 18}$ 

$\text{Sum of 30 numbers = 540}$

If each observation is increased by $2$, then new mean will be

$\Rightarrow \left( 2\times 30 \right)+540$

$\Rightarrow 60+540$

$\Rightarrow 600$

Therefore,

$Mean=\dfrac{600}{30}$

$Mean=20$

9. Find the mean of $30$ numbers where, the mean of $10$ of them is $12$ and the mean of remaining $20$ is $9$.

Ans: From the given data, we have

Total numbers $=30$

Mean of $10$ numbers is $=12$

Therefore,

$12=\dfrac{\sum\limits_{i=1}^{n}{{{X}_{i}}}}{10}$

$\sum{{{X}_{i}}}=12\times 10=120..........(1)$

Mean of $20$ numbers is $=9$

Therefore,

\[9=\dfrac{\sum{{{X}_{i}}}}{20}\]

\[9\times 10\sum\limits_{i=1}^{n}{{{X}_{i}}}\]

\[180=\sum{{{X}_{i}}}..........(2)\]

By adding eqn. \[(1)\] and \[(2)\], we get

Mean of $30$ numbers 

$=\dfrac{120+180}{30}$

$=\dfrac{300}{30}$

$=10$

Therefore, mean of $30$ numbers is $10$.

Very Short Answer Questions (1 Mark)

1. $\sum{{{f}_{i}}}=15,\sum{{{f}_{i}}}{{x}_{i}}=3p+36$ and mean of any distribution is $3$, then $p=$

1. $2$

2. $3$

3. $4$

4. $5$

Ans: b) $3$

\[\text{Mean}=\dfrac{\sum{{{\text{f}}_{\text{i}}}{{\text{x}}_{\text{i}}}}}{\sum{{{\text{f}}_{\text{i}}}}}\]

Therefore,

\[\text{3}=\dfrac{3p+36}{15}\]

\[45=3p+36\]

\[3p=45-36\]

\[3p=9\]

Therefore,

\[p=3\]

2. For what value of $x$, the mode of the following data is $8$: \[4,\text{ }5,\text{ }6,\text{ }8,\text{ }5,\text{ }4,\text{ }8,\text{ }5,\text{ }6,\text{ }x,\text{ }8\]

1. $5$

2. $6$

3. $8$

4. $4$

Ans: c) $8$

Mode is the number that appears the most times.

Here, $8$ is mode that means it occurs maximum times than others in above data and hence, value of $x$ will be $8$.

3. The numbers are arranged in ascending order. If their median is $25$ , then 

\[x\text{ }=\text{5, 7, 10, 12, 2x-8, 2x+10, 35, 41, 42, 50}\]. Find value of x

1. $10$

2. $11$

3. $12$

4. $9$

Ans: c) $12$

Here, No. of observations = $10$

$Median=\dfrac{{{\dfrac{n}{2}}^{th}}term+{{\left( \dfrac{n}{2}+1 \right)}^{th}}term}{2}$

Therefore,

$25=\dfrac{{{\dfrac{10}{2}}^{th}}term+{{\left( \dfrac{10}{2}+1 \right)}^{th}}term}{2}$

$25=\dfrac{{{5}^{th}}term+{{6}^{th}}term}{2}$

$25=\dfrac{2x-8+2x+10}{2}$

$25=2x+1$

$x=12$

4. The median for the following frequency distribution is

a). $6$

b). $5$

c). $4$

d). $7$

Ans: a) $6$

Here, No. of observations = $7$

$Median={{\left( \dfrac{n+1}{2} \right)}^{th}}term$

Therefore,

$Median={{\left( \dfrac{7+1}{2} \right)}^{th}}term$

$Median={{4}^{th}}term$

$Median={{\left( \dfrac{n+1}{2} \right)}^{th}}term$

5. In the formula $\overline{x}=a+h\left( \dfrac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right]$, for finding the mean of grouped frequency

distribution, then \[{{u}_{i}}\text{ }=\]

1. $\dfrac{{{x}_{i}}+a}{h}$

2. $h({{x}_{i}}-a)$

3. $\dfrac{{{x}_{i}}-a}{h}$

4. $\dfrac{a-{{x}_{i}}}{h}$

Ans: c) $\dfrac{{{x}_{i}}-a}{h}$

The formula for calculating the calculated mean from a grouped frequency table is

$\overline{x}=a+\left( \dfrac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}}\times h \right)$

Where, 

$a$ is assumed mean and

$h$ is size of class interval

\[{{f}_{i}}\] is frequency

\[{{u}_{i}}=\dfrac{{{x}_{i}}-a}{h}\]

Hence, Option $\text{c) }\dfrac{{{\text{x}}_{\text{i}}}\text{- a}}{\text{h}}$ is correct.

6. While computing mean of grouped data, we assume that the frequencies are

1. evenly distributed over all the class

2. centered at the class marks of the class

3. centered the upper limits of the class

4. centered the lower limits of the class

Ans: b) Centered at the class marks of the class

The frequencies are centered at the class marks of the classes when computing the mean of grouped data.

7. If \[\sum{{{f}_{i}}}=17,\sum{{{f}_{i}}}{{x}_{i}}=4P+63\] and mean \[=\text{ }7\], then $P=$

1. \[12\]

2. \[13\]

3. \[14\]

4. \[15\]

Ans: c) 14

\[\text{Mean}=\dfrac{\sum{{{\text{f}}_{\text{i}}}{{\text{x}}_{\text{i}}}}}{\sum{{{\text{f}}_{\text{i}}}}}\]

\[\text{7}=\dfrac{4P+63}{17}\]

\[P=\dfrac{119-56}{4}\]

\[P=14\]

8. If the value of mean and mode are respectively \[30\] and \[15\] , then find median.

1. \[22.5\]

2. \[24.5\]

3. \[25\]

4. \[26\]

Ans: c) \[25\]

Mode can be calculated as;

\[\text{Mode = 3 median -- 2 mean}\]

\[\text{Median = }\dfrac{\text{Mode + 2 Mean}}{3}\]

\[\text{Median = }\dfrac{\text{15 + }\left( 2\times 30 \right)}{3}\]

\[\text{Median = }25\]

9. The following is a list of a bowler's wickets in \[10\] cricket matches.

\[2,6,4,5,0,2,1,3,2,3\] Determine the mode of the data.

1. \[1\]

2. \[4\]

3. \[2\]

4. \[3\]

Ans: c) \[2\]

As per given, the number of wickets taken by a bowler in \[10\] cricket matches are as follows;

\[2,6,4,5,0,2,1,3,2,3\]

In the maximum number of matches, the bowler takes two wickets.

As a result, the data's mode is \[2\] .

10. Find the mean of the following data is

1. \[76\]

2. \[77\]

3. \[78\]

4. \[80\]

Ans: c) \[78\]

Now, we can find out mean by using the following formulae,

\[\text{Mean}=\dfrac{\sum{{{\text{f}}_{\text{i}}}{{\text{x}}_{\text{i}}}}}{\sum{{{\text{f}}_{\text{i}}}}}\]

\[\text{Mean}=\dfrac{3900}{50}\]

Therefore,

\[\text{Mean}=78\]

Therefore, the mean of the given above data is \[78\].

11. Construction of a cumulative frequency table is useful in determining the

1. Mean

2. Median

3. Mode

4. all these conditions

Ans: b) Median

The median of the series can be determined using a cumulative frequency table.

12. The heights of \[60\] students in a class are distributed in the below table. 

The sum of the lower limit of the modal class and upper limit of the median class is

1. \[310\]

2. \[315\]

3. \[320\]

4. \[330\]

Ans:  b) \[315\]

13. Choose the correct answer from the given four options in the formula

$\overline{x}=a+\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}$

For finding the mean of grouped data ${{d}_{i}}$’s are deviations from a of

1. lower limits of the classes

2. Upper limits of the classes

3. Mid points of the classes

4. Frequencies of the class marks

Ans:  c) Mid points of the classes

14. If mean of the distribution is \[7.5\], then find \[P\]

1. \[2\]

2. \[4\]

3. \[3\]

4. \[6\]

Ans:  c) \[3\]

15. A shoe shop in Agra had sold hundred pairs of shoes of particular brand in a certain day with the following distribution.

Find mode of the destitution.

1. \[20\]

2. \[45\]

3. \[1\]

4. \[3\]

Ans:  b) \[45\]

16. If the mode of a data is \[45\] and mean is \[27\], then median is

1. \[30\]

2. \[27\]

3. \[33\]

4. None of these

No. of pairs sold \[1,4,3,20,45,25,2\]

Find mode of the destitution.

1. \[20\]

2. \[45\]

3. \[1\] 

4. \[3\]

Ans: c) \[1\]

17. If ${{x}_{i}}$’s are the mid-points of the class intervals of grouped data, ${{f}_{i}}$’s are the corresponding frequency and is the mean $\overline{x}$, then $\sum{\left( {{f}_{i}}{{x}_{i}}-\overline{x} \right)}$is equal to

1. \[0\]

2. \[-1\]

3. \[1\]

4. \[2\]

Ans:  a) \[0\]

18. Mode of the following data is

1. \[65\]

2. \[66\]

3. \[75\]

4. \[70\]

Ans:  d) \[70\]

19. Median of the following data is

1. \[1000\]

2. \[1100\]

3. \[1200\]

4. \[1150\]

Ans:  c) \[1200\]

20. If the median of the distribution is \[28.5\], find the value of $x$.

1. \[8\]

2. \[10\]

3. \[4\]

4. \[9\]

Ans:  a) \[8\]

Short Answer Questions (2 Marks)

1. In a class of $40$ students, the following data shows the number of boys of a specific age. Determine the mean age of the students.

Ans: We have

Mean can be calculated as;

Mean,

$\overline{x}=\dfrac{\sum{fx}}{\sum{f}}$

 \[\text{=}\dfrac{\text{698}}{\text{40}}\]

Therefore,

\[\overline{\text{x}}\text{ = 17}\text{.45 years}\]

2. For the following grouped frequency distribution, determine the mode.

Ans: Here, the maximum frequency \[=\text{ }23\] and lies into the class $12-15$.

Therefore,

Modal class is $12-15$

We have,

$l=12,n=3,{{f}_{1}}=23,{{f}_{0}}=10,{{f}_{2}}=21$

Mode can be calculated as,

${{M}_{0}}=l+h\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}$

$=12+3\dfrac{23-10}{2\times 23-10-21}$

\[=12+3\times \dfrac{13}{46-31}\]

\[=12+\dfrac{39}{15}\]

\[=12+\dfrac{13}{5}\]

\[=12+2.6\]

 \[=14.6\]

Therefore, mode frequency is \[14.6\].

3. Construct the cumulative frequency distribution of the following distribution:

Ans: The required cumulative frequency (C.F.) distribution of the given distribution is given below:

4. The median and mode of a distribution are $21.2$ and $21.4$ respectively, determine its mean.

Ans: Mode can be calculated as,

$Mode=3Median-2Mean$

Therefore,

$Mean=Mode+\dfrac{3}{2}\left( Median-Mode \right)$

$=21.4+\dfrac{3}{2}(21.2-21.4)$

$=21.4+\dfrac{3}{2}(-0.2)$

$=21.4-0.3$

Therefore, $Mean=21.1$

5. The marks distribution of $30$ students in a mathematics examination are given below

Ans: Here, the maximum frequency \[=\text{ }7\]and lies into to the class $40-55$.

Therefore, the modal class is $40-55$

We have, 

$l=40,h=15,{{f}_{1}}=7,{{f}_{0}}=3,{{f}_{2}}=6$

Mode ${{M}_{0}}$is given by

\[{{M}_{0}}=l+h\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\]

\[=40+\dfrac{15(7-3)}{2(7)-3-6}\]

\[=40+\dfrac{15\times 4}{5}\]

 \[=40+12\]

\[{{M}_{0}}=52\]

Therefore, Mode marks is $52$.

6. Find the mode of this data.

Construct the cumulative frequency distribution of following distribution:

Ans: The required cumulative frequency distribution of the above given distribution is as follows:

7. If the values of mean and mode are $30$ and $15$ respectively, then find out its median.

1. $22.5$

2. $24.5$

3. $25$

4. $26$

Ans: Mode can be calculated as,

$Mode=3Median-2Mean$

Therefore,

$Median=Mode+\dfrac{2}{3}(Mean-Mode)$

$=15+\dfrac{2}{3}(30-15)$

$=15+\dfrac{2}{3}\times 15$

 \[=15+10\]

Therefore, \[Median=25\]

8. If the mean of the following data is $18.75$. Find the value of $P$.

Ans: We have,

Mean can be calculated as,

Mean,

\[\overline{x}=\dfrac{\sum{{{f}_{i}}}{{x}_{i}}}{\sum{{{f}_{i}}}}\]

Therefore, $18.75=\dfrac{406+7P}{32}$

$\Rightarrow 460+7P=\dfrac{32\times 1875}{1000}$

\[\Rightarrow 460+7P=600\]

\[\Rightarrow 7P=600-460\]

\[\Rightarrow 7P=140\]

Therefore,

\[\Rightarrow P=20\]

9. Compute the mean of the following data.

Ans: We have,

Mean is given by,

Mean, \[\overline{x}=\dfrac{\sum{{{f}_{i}}}{{x}_{i}}}{\sum{{{f}_{i}}}}\]

\[=\dfrac{1900}{50}\]

\[=38\]

Hence, Mean $=38$.

10. The following data gives the information observed life times (in hours) of $225$ electrical components. Determine the modal life times of the components.

Ans: We have the maximum frequency is $61$ and it lies into the class $60-80$

Hence, Modal class $=60-80$

We have, 

$l=60,h=20,{{f}_{1}}=61,{{f}_{0}}=52,{{f}_{2}}=38$

The mode is given by

${{M}_{0}}=l+h\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}$

$=60+20\dfrac{61-52}{2(61)-52-38}$

$=60+20\dfrac{9}{122-90}$

$=60+\dfrac{20\times 9}{32}$

$=60+\dfrac{45}{8}$

$=60+5.625$

$=65.625$

Therefore, modal life times $\text{= 65}\text{.625 hrs}$.

11. Construct the cumulative frequency distribution of the following distribution:

Ans: The required cumulative frequency distribution of the given distribution is given as follows:

12. Calculate the median from the following data:

Ans: We have,

Here,

$N=60$

Therefore $\dfrac{N}{2}=30$ that lies in the class $30-60$

Therefore, Median Class \[=30-60\]

Size of interval, \[h=60-30=30\]

Lower limit of Modal class, \[l=30\]

\[CF=20\]

\[\text{Median}=l+\dfrac{\dfrac{N}{2}-CF}{f}\times h\]

Therefore,\[\text{Median}=30+\dfrac{30-20}{30}\times 30\]

Therefore,\[\text{Median}=30+10\]

\[\text{Median}=40\]

Therefore, the median is \[40\].

13. Find the mean of the following data:

Ans: We have,

Mean can be calculated as,

Mean,

\[\overline{x}=\dfrac{\sum{{{f}_{i}}}{{x}_{i}}}{\sum{{{f}_{i}}}}\]

 \[=\dfrac{625}{25}\]

\[=25\]

14. The following frequency chart for the number of family members in a household was derived from a survey conducted by a group of students on $20$ families in a neighborhood. Compute the mode. 

Ans: Here, the maximum frequency is $8$ and it lies to the class $3-5$.

Hence, Modal class $=3-5$

We have,

$l=3,h=2,{{f}_{1}}=8,{{f}_{0}}=7,{{f}_{2}}=2$

Mode can be given as;

${{M}_{0}}=l+h\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}$

$=3+2\dfrac{(8-7)}{2(8)-7-2}$

$=3+2\dfrac{1}{7}$

$=3+\dfrac{2}{7}$

Therefore,

$Mode=3.286$

15. Construct the cumulative frequency distribution of the following distribution:

Ans: The required cumulative frequency distribution of the given distribution is given below:

16. The values of mean and median are $26.4$ and $27.2$ respectively, find the value of mode.

Ans:Mode can be calculated as,

\[Mode\text{ }=\text{ }3\text{ }median\text{ }-2\text{ }mean\]

\[=\text{ }3\left( 27.2 \right)\text{ }\text{ }2\left( 26.4 \right)\]

\[=\text{ }81.6\text{ }\text{ }52.8\text{ }=\text{ }28.8\]

Therefore,

\[Mode\text{ }=\text{ }28.8\]

17. The results of $30$ students from a particular school's class $X$ in a $100$ -point Mathematics paper are shown in the table below. Calculate the mean of the students' marks.

Ans:

Mean,

$\overline{x}=\dfrac{\sum{{{f}_{i}}}{{x}_{i}}}{\sum{{{f}_{i}}}}$

$=\dfrac{1779}{30}$

$=59.3$

Therefore, mean $\overline{x}=59.3$

18. A student recorded the number of cars passing via a particular point on a road for $100$ periods of $3$ minutes each, which he then summarized in the table below. Determine the data's mode.

Ans: Here, the maximum frequency is $20$ and it corresponds to the class $40-50$

Hence, Modal class $=40-50$

We have, 

$l=40,h=10,{{f}_{1}}=20,{{f}_{0}}=12,{{f}_{2}}=11$

Mode is given by,

${{M}_{0}}=l+h\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}$

$=40+10\left( \dfrac{20-12}{2(20)-12-11} \right)$

$=40+\dfrac{80}{17}$

$=40+4.705$

Therefore,

$Mode=44.71$

.

19. Construct the cumulative frequency distribution of the following distribution:

Ans: The required accumulative frequency distribution of the given distribution is given below:

20. The values of mean and median are $53.6$ and $55.81$ respectively, find out the value of mode.

Ans: We know that

\[\text{Mode = 3 Median - 2 Mean}\]

Therefore,

$Mode=3(55.81)-2(53.6)$

$=167.43-107.2$

Therefore,

$Mode=60.23$

Short Answer Questions (3 Marks)

1. The following table shows the weekly wages drawn by number of workers in a factory, determine the median.

Ans: We have,

Here,

$N=188$

Therefore $\dfrac{N}{2}=94$ that lies in the class \[200-300\]

Therefore, Median Class \[=200-300\]

Size of interval, \[h=300-200=100\]

Lower limit of Modal class, \[l=200\]

\[CF=79\]

\[\text{Median}=l+\dfrac{\dfrac{N}{2}-CF}{f}\times h\]

Therefore,

\[\text{Median}=200+\dfrac{94-79}{34}\times 100\]

Therefore,

\[\text{Median}=200+44.12\]

\[\text{Median}=244.12\]

Therefore, the median is \[244.12\].

2. Determine the median of the following data:

Ans: First, we need to convert cumulating series into simple series.

Here,

$N=100$

Therefore $\dfrac{N}{2}=50$ that lies in the class \[70-90\]

Therefore, Median Class \[=70-90\]

Size of interval, \[h=90-70=20\]

Lower limit of Modal class, \[l=70\]

\[CF=43\]

\[\text{Median}=l+\dfrac{\dfrac{N}{2}-CF}{f}\times h\]

Therefore,

\[\text{Median}=70+\dfrac{50-43}{22}\times 20\]

Therefore,

\[\text{Median}=70+6.36\]

\[\text{Median}=76.36\]

Therefore, the median is \[76.36\].

3. Find the median of the following data.

Ans: First, we need to calculate the simple frequencies.

Here,

$N=150$

Therefore $\dfrac{N}{2}=75$ that lies in the class \[110-120\]

Therefore, Median Class \[=110-120\]

Size of interval, \[h=120-110=10\]

Lower limit of Modal class, \[l=110\]

\[CF=45\]

\[\text{Median}=l+\dfrac{\dfrac{N}{2}-CF}{f}\times h\]

Therefore, \[\text{Median}=110+\dfrac{75-45}{45}\times 10\]

Therefore,\[\text{Median}=110+6.67\]

\[\text{Median}=116.67\]

Therefore, the median is \[116.67\].

4. Draw a less than Ogive for the following frequency distribution.

Ans: We have,

The required Ogive which is as follows:

(Image will be uploaded soon)

5. Find the mean age in years from the frequency distribution given below:

Ans: We have,

Here, Assumed mean,  \[a=32\], \[h=5\]

Therefore,

Mean,

$\overline{x}=a+\dfrac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}}\times h$

$=32\dfrac{-37\times 5}{63}$

$=32\dfrac{-185}{63}$

$=32-2.94$

Therefore,

$\overline{x}=\text{29}\text{.06 years}$

Therefore, mean age is $\text{29}\text{.06 years}$.

6. Find the median of the following frequency distribution:

Ans: We have,

Here,

$N=44$

Therefore $\dfrac{N}{2}=22$ that lies in the class \[400-500\]

Therefore, Median Class \[=400-500\]

Size of interval, \[h=500-400=100\]

Lower limit of Modal class, \[l=400\]

\[CF=8\]

\[\text{Median}=l+\dfrac{\dfrac{N}{2}-CF}{f}\times h\]

Therefore,\[\text{Median}=400+\dfrac{22-8}{20}\times 100\]

Therefore,\[\text{Median}=400+70\]

\[\text{Median}=470\]

Therefore, the median is \[470\].

7. The following data shows production yield per hectare of wheat of \[100\] farms of village:

Change the distribution to a more than type distribution and draw its Ogive.

Ans: We will consider here more than type Ogive

Now by plotting the points \[\left( 50,100 \right)\], \[\left( 55,98 \right)\], \[\left( 60,90 \right)\] , \[\left( 65,78 \right)\] , \[\left( 70,54 \right)\] , \[\left( 75,16 \right)\] , the Ogive formed is as follows;

(Image will be uploaded soon)

8. The A.M of the following distribution is \[47\]. Determine the value of \[P\] .

Ans: We have,

Mean can be calculated as,

Mean, $\overline{x}=\dfrac{\sum{{{f}_{i}}}{{x}_{i}}}{\sum{{{f}_{i}}}}$

Therefore,

$\Rightarrow 47=\dfrac{1980+70P}{48+P}$

$\Rightarrow 2256+47P=1980+70P$

$\Rightarrow 70P-47P=2250-1980$

$\Rightarrow 23P=276$

$\Rightarrow P=\dfrac{276}{23}$

$\Rightarrow P=12$

Hence,

Value of $P$ is $12$.

9. A doctor examined thirty ladies in a hospital, and the number of heart beats per minute was recorded and reported as follows: 

Determine the mean heartbeats per minute for these ladies using the most appropriate approach.

Ans: Here, assumed mean, \[a=75.5\]. 

We have,

By using Step deviation method, Mean can be calculated as,

Mean \[\overline{x}=a+\dfrac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}}\times h\]

$=75.5+\dfrac{4}{30}\times 3$

$=75.5+0.4$

Therefore, $\overline{x}=75.9$

10. Following distribution gives the marks obtained by a class of $100$ students:

Change the distribution to less than type distribution and draw its Ogive.

Ans: We will consider here less than type Ogive.

Now by plotting the points \[\left( 20,10 \right)\], \[\left( 30,25 \right)\], \[\left( 40,55 \right)\] , \[\left( 50,87 \right)\] , \[\left( 60,95 \right)\] , \[\left( 70,100 \right)\] , the Ogive formed is as follows;

(Image will be uploaded soon)

11. The daily pocket allowances given to the children of a multistory building are shown in the table below. The mean amount of pocket money is $\text{Rs}\text{.18}$ . Find out the missing frequency.

Ans: Let’s consider \[f\]be the missing frequency, we have

Here, assumed mean, \[a=18\], and \[h=2\]

Mean can be calculated as;

Mean,

\[\overline{x}=a+\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\times h\]

\[\Rightarrow 18=18+\dfrac{\left( f-8 \right)}{40+f}\times 2\]

\[\Rightarrow 0=f-8\]

Therefore,

 \[f=8\]

Thus, missing frequency is \[8\].

12. The following data was acquired from a survey of  \[51\] girls in Class \[X\] of a school on their heights (in cm). Calculate the median height.

Ans:

We have,

Here,

$N=51$

Therefore $\dfrac{N}{2}=25.2$ that lies in the class  \[145-150\]

Therefore, Median Class \[=145-150\]

Size of interval, \[h=150-145=5\]

Lower limit of Modal class \[l=145\]

\[CF=11\]

\[\text{Median}=l+\dfrac{\dfrac{N}{2}-CF}{f}\times h\]

Therefore,

\[\text{Median}=145+\dfrac{25.2-11}{18}\times 5\]

Therefore,

\[\text{Median}=145+\dfrac{72.5}{18}\]

Therefore,

\[\text{Median}=149.03\]

Therefore, the median height of girls is \[149.03\].

13. For the following distribution, find the mean.

Ans: We have,

Now, we can find out mean by using the following formulae,

Mean:

\[\text{Mean}=\dfrac{\sum{{{\text{f}}_{\text{i}}}{{\text{x}}_{\text{i}}}}}{\sum{{{\text{f}}_{\text{i}}}}}\]

\[\text{Mean}=\dfrac{1108}{66}\]

Therefore,

\[\text{Mean}=16.79\]

Therefore, the mean of the given above data is \[16.79\].

14. The following table shows the proportion of marks earned by 100 students in an examination:

Calculate the median percentage of marks.

Ans:

Here,

$N=100$

Therefore $\dfrac{N}{2}=50$ that lies in the class \[45-50\]

Therefore, Median Class\[=45-50\]

Size of interval, \[h=50-45=5\]

Lower limit of Modal class \[l=45\]

\[CF=48\]

\[\text{Median}=l+\dfrac{\dfrac{N}{2}-CF}{f}\times h\]

Therefore,

\[\text{Median}=45+\dfrac{50-48}{23}\times 5\]

Therefore,

\[\text{Median}=45+\dfrac{10}{23}\]

\[\text{Median}=45.44\]

Therefore, the median percentage of marks is \[45.44\].

15. For the following frequency distribution, draw a less than Ogive.

Ans: We have,

We acquire the needed Ogive by joining these points with a free hand curve, which is as follows:

(Image will be uploaded soon)

16. The following frequency distribution has an A.M of \[53\] . Find the value of $P$ .

Ans: We have,

Mean can be calculated as;

Mean $\overline{x}=\dfrac{\sum{{{f}_{i}}}{{x}_{i}}}{\sum{{{f}_{i}}}}$

$\Rightarrow 53=\dfrac{3340+70P}{72+P}$

$\Rightarrow 3340+70P=3816+53P$

$\Rightarrow 17P=3816-3340$

$\Rightarrow P=\dfrac{476}{17}$

$\therefore P=28$

Therefore, value of $P$ is $28$.

Long Answer Questions (4 Marks)

1. In the following distribution, locate the median mean and mode.

Ans: We have,