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NCERT Solutions for Class 11 Maths Chapter 15: Statistics - Exercise 15.1

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NCERT Solutions for Class 11 Maths Chapter 15 Statistics

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.1 (Ex 15.1) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 15 Statistics Exercise 15.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 11

Subject:

Class 11 Maths

Chapter Name:

Chapter 15 - Statistics

Exercise:

Exercise - 15.1

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

Available Materials:

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Other Materials

  • Important Questions

  • Revision Notes

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Access NCERT Solutions for Class 11 Maths Chapter 15 –Statistics

Exercise 15.1

1. Calculate the mean deviation about mean for the numbers 

$\mathbf{4},\mathbf{7},\mathbf{8},\mathbf{9},\mathbf{10},\mathbf{12},\mathbf{13},\mathbf{17}$.

Ans.

We are provided the numbers

$4,7,8,9,10,12,13,17$.

Therefore, mean of the numbers,

 $ \overline{x}=\dfrac{4+7+8+9+10+12+13+17}{8} $

$ =\dfrac{80}{8} $

$ =10 $

The deviations for the given numbers with the mean $\overline{x}=10$, 

that is ${{x}_{i}}-\overline{x}$, for $i=1,2,...,8$ are 

$-6,-3,-2,-1,\,\,0,\,\,2,\,\,3,\,\,7$.

Also, the absolute values of the obtained deviations, that is $\left| {{x}_{i}}-x \right|$ are

$6,3,2,1,0,2,3,7$.

Thus, the mean deviation about the mean $\overline{x}$ for the given numbers is

$ M.D.\left( \overline{x} \right)=\dfrac{\sum\limits_{i=1}^{8}{\left| {{x}_{i}}-overline{x} \right|}}{8}\\ $

 $ =\dfrac{6+3+2+1+0+2+3+7}{8}\\ $

$ =\dfrac{24}{8}\\ $

$ =3\\ $


2. Calculate the mean deviation about the mean for the following numbers.

$\mathbf{38},\mathbf{70},\mathbf{48},\mathbf{40},\mathbf{42},\mathbf{55},\mathbf{63},\mathbf{46},\mathbf{54},\mathbf{44}$

Ans.

We are provided that numbers

$\text{38,70,48,40,42,55,63,46,54,44}$.

Therefore, the mean of the numbers,

$ \overline{x}=\dfrac{38+70+48+40+42+55+63+46+54+44}{10}\\ $

$ =\dfrac{500}{10}\\ $

$ =50\\ $

Now, the deviations for the given numbers with the mean $\overline{x}=50$, that is

${{x}_{i}}-x$, for $i=1,2,3,...,10$ are 

$-12,20,-2,-10,-8,5,13,-4,4,-6$.

So, the absolute values of the obtained deviations for the given numbers, that is $\left| {{x}_{i}}-x \right|$ are

$12,20,2,10,8,5,13,4,4,6$.

Thus, the mean deviation about the mean $\overline{x}$ for the given numbers is

$M.D.\left( \overline{x} \right)=\dfrac{\sum\limits_{i=1}^{10}{\left| {{x}_{i}}-x \right|}}{10}$

$=\dfrac{12+20+2+10+8+5+13+4+4+6}{10}$

$=\dfrac{84}{10}$

$=8.4$


3. Calculate the mean deviation about the median for the following numbers.

$\mathbf{13},\mathbf{17},\mathbf{16},\mathbf{14},\mathbf{11},\mathbf{13},\mathbf{10},\mathbf{16},\mathbf{11},\mathbf{18},\mathbf{12},\mathbf{17}$.

Ans.

We are provided the numbers

$\text{13,17,16,14,11,13,10,16,11,18,12,17}$.

There is total $12$ numbers, and it is even.

Now, arranging the numbers in the ascending order, gives

$10,11,11,12,13,13,14,16,16,17,17,18$.

It is known that, 

Median, $M=\dfrac{{{\left( \dfrac{n}{2} \right)}^{th}}\,\,number+{{\left( \dfrac{n}{2}+1 \right)}^{th}}number}{2}$, when $n$ is even.

Therefore, the median

$M=\dfrac{{{\left( \dfrac{12}{2} \right)}^{th}}number+{{\left( \dfrac{12}{2}+1 \right)}^{th}}number}{2}$

$=\dfrac{{{6}^{th}}\,number+{{7}^{th}}\,\,number}{2}$

$=\dfrac{13+14}{2}$

$=\dfrac{27}{2}$

$=13.5$

Now, the deviations about the median $M$ for the given numbers, that is ${{x}_{i}}-M$, for $i=1,2,3,...,12$ are

$-3.5,-2.5,-2.5,-1.5,-0.5,-0.5,\,0.5,\,2.5,\,3.5,\,3.5,\,3.5,\,4.5$

So, the absolute values of the resulted deviations, that is $\left| {{x}_{i}}-M \right|$, for $i=1,2,3,...,12$ are

$3.5,\,2.5,\,2.5,\,1.5,\,0.5,\,0.5,\,0.5,\,2.5,\,2.5,\,3.5,\,3.5,\,4.5\,$.

Thus, the mean deviation about the median $M$ for the given numbers,

$M.D.\left( M \right)=\dfrac{\sum\limits_{i=1}^{12}{\left| {{x}_{i}}-M \right|}}{12}$

$=\dfrac{3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+4.5}{12}$

$=\dfrac{28}{12}$

$=2.33$


4. Calculate the mean deviation about the median for the following numbers.

$\mathbf{36},\,\mathbf{72},\,\mathbf{46},\,\mathbf{42},\,\mathbf{60},\,\mathbf{45},\,\mathbf{53},\,\mathbf{46},\,\mathbf{51},\,\mathbf{49}$.

Ans.

We are provided the numbers

$\text{36,}\,\text{72,}\,\text{46,}\,\text{42,}\,\text{60,}\,\text{45,}\,\text{53,}\,\text{46,}\,\text{51,}\,\text{49}$.

There is total $10$ numbers, and it is even.

Now, arranging the numbers in the ascending order, gives

$36,\,42,\,45,\,46,\,46,\,49,\,51,\,53,\,60,\,72$.

It is known that, 

Median, $M=\dfrac{{{\left( \dfrac{n}{2} \right)}^{th}}\,\,number+{{\left( \dfrac{n}{2}+1 \right)}^{th}}number}{2}$, when $n$ is even.

Therefore, the median

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

$=\dfrac{{{5}^{th}}\,number+{{6}^{th}}\,\,number}{2}$

$=\dfrac{46+49}{2}$

$=\dfrac{95}{2}$

$=47.5$

Now, the deviations about the median $M$ for the given numbers, that is ${{x}_{i}}-M$, for $i=1,2,3,...,10$ are

$-11.5,\,-5.5,\,-2.5,\,-1.5,\,-1.5,\,1.5,\,3.5,\,5.5,\,12.5,\,24.5$

So, the absolute values of the resulted deviations, that is $\left| {{x}_{i}}-M \right|$, for $i=1,2,3,...,10$ are

$11.5,\,5.5,\,2.5,\,1.5,\,1.5,\,1.5,\,3.5,\,5.5,\,12.5,\,24.5$.

Thus, the mean deviation about the median $M$ for the given numbers,

$M.D.\left( M \right)=\dfrac{\sum\limits_{i=1}^{10}{\left| {{x}_{i}}-M \right|}}{10}$

$=\dfrac{11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5}{10}$

$=\dfrac{70}{10}$

$=7$.


5. Use the following data to calculate the mean deviation about the mean.

${{\mathbf{x}}_{\mathbf{i}}}$

$\mathbf{5}$

$\mathbf{10}$

$\mathbf{15}$

$\mathbf{20}$

$\mathbf{25}$

${{\mathbf{f}}_{\mathbf{i}}}$

$\mathbf{7}$

$\mathbf{4}$

$\mathbf{6}$

$\mathbf{3}$

$\mathbf{5}$


Ans.

Consider the following table of data.

${{x}_{i}}$

${{f}_{i}}$

${{f}_{i}}{{x}_{i}}$

$\left| {{x}_{i}}-\overline{x} \right|$

${{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|$

$5$

$7$

$35$

$9$

$63$

$10$

$4$

$40$

$4$

$16$

$15$

$6$

$90$

$1$

$6$

$20$

$3$

$60$

$6$

$18$

$25$

$5$

$125$

$11$

$55$

Total

$\sum{{{f}_{i}}=}25$

$\sum{{{f}_{i}}{{x}_{i}}=}350$


$\sum{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|}=158$


Thus, from the above table we have,

$N=\sum\limits_{i=1}^{5}{{{f}_{i}}}=25$,

$\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}=350}$

Therefore, the mean, 

$ \overline{x}=\dfrac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}}\\ $

$  =\dfrac{1}{25}\times 350\\ $

$ =14\\ $

Hence, the mean deviation about the mean is given by

$ M.D\left( \overline{x} \right)=\dfrac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|}\\ $

$ =\dfrac{1}{25}\times 158\\ $

$  =6.32\\ $


6. Use the following data to calculate the mean deviation about the mean.

${{\mathbf{x}}_{\mathbf{i}}}$

$\mathbf{10}$

$\mathbf{30}$

$\mathbf{50}$

$\mathbf{70}$

$\mathbf{90}$

${{\mathbf{f}}_{\mathbf{i}}}$

$\mathbf{4}$

$\mathbf{24}$

$\mathbf{28}$

$\mathbf{16}$

$\mathbf{8}$


Ans.

Consider the following table of data.

${{x}_{i}}$

${{f}_{i}}$

${{f}_{i}}{{x}_{i}}$

$\left| {{x}_{i}}-\overline{x} \right|$

${{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|$

$10$

$4$

$40$

$40$

$160$

$30$

$24$

$720$

$20$

$480$

$50$

$28$

$1400$

$0$

$0$

$70$

$16$

$1120$

$20$

$320$

$90$

$8$

\[720\]

$40$

$320$

Total

$\sum{{{f}_{i}}=}80$

$\sum{{{f}_{i}}{{x}_{i}}=}4000$


$\sum{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|}=1280$


Thus, from the above table we have,

$N=\sum\limits_{i=1}^{5}{{{f}_{i}}}=80$,

$\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}=}4000$

Therefore, the mean, 

$ \overline{x}=\dfrac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}{{x}_{i}}}\\ $

$ =\dfrac{1}{80}\times 4000\\ $

 $ =50\\ $

Hence, the mean deviation about the mean is given by

$ M.D\left( \overline{x} \right)=\dfrac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|}\\ $

$ =\dfrac{1}{80}\times 1280\\ $

$ =16\\ $.


7. Use the following data to calculate the mean deviation about the median.

${{\mathbf{x}}_{\mathbf{i}}}$

$\mathbf{5}$

$\mathbf{7}$

$\mathbf{9}$

$\mathbf{10}$

$\mathbf{12}$

$\mathbf{15}$

${{\mathbf{f}}_{\mathbf{i}}}$

$\mathbf{8}$

$\mathbf{6}$

$\mathbf{2}$

$\mathbf{2}$

$\mathbf{2}$

$\mathbf{6}$


Ans.

Notice that the given data is already in ascending order.

Consider the following table that shows the cumulative frequencies of the provided data.

${{x}_{i}}$

${{f}_{i}}$

$c.f.$

$5$

$8$

$8$

$7$

$6$

$14$

$9$

$2$

$16$

$10$

$2$

$18$

$12$

$2$

$20$

$15$

$6$

$26$


Thus, from the above table we have,

$N=26$, that is an even number.

Therefore, the median can be obtained by calculating the mean of the ${{13}^{th}}$ and ${{14}^{th}}$ observations and both the observations belongs to the cumulative frequency $14$.

Therefore,

Median,

 $ M=\dfrac{{{13}^{th}}\,\,observation+{{14}^{th}}\,observation}{2}\\ $

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

$ =7\\ $

Now, the following table shows the absolute values of the deviations about the median, that is $\left| {{x}_{i}}-M \right|$.

$\left| {{x}_{i}}-M \right|$

$2$

$0$

$2$

$3$

$5$

$8$

Total

${{f}_{i}}$

$8$

$6$

$2$

$2$

$2$

$6$

$\sum{{{f}_{i}}=26}$

${{f}_{i}}\left| {{x}_{i}}-M \right|$

$16$

$0$

$4$

$6$

$10$

$48$

$\sum{{{f}_{i}}\left| {{x}_{i}}-M \right|=84}$


Thus, from the above table, we have

$\sum\limits_{i=1}^{6}{{{f}_{i}}}=26$ and $\sum\limits_{i=1}^{6}{{{f}_{i}}\left| {{x}_{i}}-M \right|}=84$.

Hence, the required mean deviation about the median,

$ M.D\left( M \right)=\dfrac{1}{N}\sum\limits_{i=1}^{6}{{{f}_{i}}\left| {{x}_{i}}-M \right|}\\ $

$ =\dfrac{1}{26}\times 84\\ $

$ =3.23\\ $


8. Use the following data to calculate the mean deviation about the median.

${{\mathbf{x}}_{\mathbf{i}}}$

$\mathbf{15}$

$\mathbf{21}$

$\mathbf{27}$

$\mathbf{30}$

$\mathbf{35}$

${{\mathbf{f}}_{\mathbf{i}}}$

$\mathbf{3}$

$\mathbf{5}$

$\mathbf{6}$

$\mathbf{7}$

$\mathbf{8}$


Ans.

Notice that the given data is already in ascending order.

Consider the following table that shows the cumulative frequencies of the provided data.


${{x}_{i}}$

${{f}_{i}}$

$c.f.$

$15$

$3$

$3$

$21$

$5$

$8$

$27$

$6$

$14$

$30$

$7$

$21$

$35$

$8$

$29$


Thus, from the above table we have,

$N=29$, that is an odd number.

Therefore, the median will be the $\dfrac{29+1}{2}=\dfrac{30}{2}={{15}^{th}}$ observation.

Notice that, the ${{15}^{th}}$ observation belongs to the cumulative frequency $21$. So, the median is the corresponding observation, that is $30$.

Now, the following table shows the absolute values of the deviations about the median, that is $\left| {{x}_{i}}-M \right|$.

$\left| {{x}_{i}}-M \right|$

$15$

$9$

$3$

$0$

$5$

Total

${{f}_{i}}$

$3$

$5$

$6$

$7$

$8$

$\sum{{{f}_{i}}=29}$

${{f}_{i}}\left| {{x}_{i}}-M \right|$

$45$

$45$

$18$

$0$

$40$

$\sum{{{f}_{i}}\left| {{x}_{i}}-M \right|=148}$


Thus, from the above table, we have

$\sum\limits_{i=1}^{5}{{{f}_{i}}}=29$ and $\sum\limits_{i=1}^{5}{{{f}_{i}}\left| {{x}_{i}}-M \right|}=148$.

Hence, the required mean deviation about the median,

  $ M.D\left( M \right)=\dfrac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}\left| {{x}_{i}}-M \right|} $

 $ =\dfrac{1}{29}\times 148 $

 $ =5.1 $

9. Use the following data to calculate the mean deviation about the mean.

Income/ day

Number of persons

$\mathbf{0}-\mathbf{100}$

$\mathbf{4}$

$\mathbf{100}-\mathbf{200}$

$\mathbf{8}$

$\mathbf{200}-\mathbf{300}$

$\mathbf{9}$

$\mathbf{300}-\mathbf{400}$

$\mathbf{10}$

$\mathbf{400}-\mathbf{500}$

$\mathbf{7}$

$\mathbf{500}-\mathbf{600}$

$\mathbf{5}$

$\mathbf{600}-\mathbf{700}$

$\mathbf{4}$

$\mathbf{700}-\mathbf{800}$

$\mathbf{3}$


Ans.

Consider the following table of data.

Income/day

Number of persons ${{f}_{i}}$

Mid-point

${{x}_{i}}$

${{f}_{i}}{{x}_{i}}$

$\left| {{x}_{i}}-\overline{x} \right|$

${{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|$

$0-100$

$4$

$50$

$200$

$308$

$1232$

$100-200$

$8$

$150$

$1200$

$208$

$1664$

$200-300$

$9$

$250$

$2250$

$108$

$972$

$300-400$

$10$

$350$

$3500$

$8$

$80$

$400-500$

$7$

$450$

$3150$

$92$

$644$

$500-600$

$5$

$550$

$2750$

$192$

$960$

$600-700$

$4$

$650$

$2600$

$292$

$1168$

$700-800$

$3$

$750$

$2250$

$392$

$1176$

Total

$\sum{{{f}_{i}}}=50$


$\sum{{{f}_{i}}{{x}_{i}}}=17900$


$7896$


Thus, from the above table we have,

\[N=\sum\limits_{i=1}^{8}{{{f}_{i}}}=50\],

$\sum\limits_{i=1}^{8}{{{f}_{i}}{{x}_{i}}=}17900$

Therefore, the mean,  

$ \overline{x}=\dfrac{1}{N}\sum\limits_{i=1}^{8}{{{f}_{i}}{{x}_{i}}} $

 $ =\dfrac{1}{50}\times 17900 $

 $ =358 $

Hence, the mean deviation about the mean is given by

$ M.D\left( \overline{x} \right)=\dfrac{1}{N}\sum\limits_{i=1}^{8}{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|}\\ $

$ =\dfrac{1}{50}\times 7896\\ $

$=157.92\\ $


10. Use the following data to calculate the mean deviation about the mean.

Height (cm)

Number of men

$\mathbf{95}-\mathbf{105}$

$\mathbf{9}$

$\mathbf{105}-\mathbf{115}$

$\mathbf{13}$

$\mathbf{115}-\mathbf{125}$

$\mathbf{26}$

$\mathbf{125}-\mathbf{135}$

$\mathbf{30}$

$\mathbf{135}-\mathbf{145}$

$\mathbf{12}$

$\mathbf{145}-\mathbf{155}$

$\mathbf{10}$


Ans.

Consider the following table of data.

Height (cm)

Number of men ${{f}_{i}}$

Mid-point

${{x}_{i}}$

${{f}_{i}}{{x}_{i}}$

$\left| {{x}_{i}}-\overline{x} \right|$

${{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|$

$95-105$

$9$

$100$

$900$

$25.3$

$227.7$

$105-115$

$13$

$110$

$1430$

$15.3$

$198.9$

$115-125$

$26$

$120$

$3120$

$5.3$

$137.8$

$125-135$

$30$

$130$

$3900$

$4.7$

$141$

$135-145$

$12$

$140$

$1680$

$14.7$

$176.4$

$145-155$

$10$

$150$

$1500$

$24.7$

$247$

Total

$\sum{{{f}_{i}}}=100$


$\sum{{{f}_{i}}{{x}_{i}}}=12530$


$1128.8$


Thus, from the above table we have,

\[N=\sum\limits_{i=1}^{6}{{{f}_{i}}}=100\],

$\sum\limits_{i=1}^{6}{{{f}_{i}}{{x}_{i}}=}12530$

Therefore, the mean,

 $\overline{x}=\dfrac{1}{N}\sum\limits_{i=1}^{6}{{{f}_{i}}{{x}_{i}}} \\ $ 

 $=\dfrac{1}{100}\times 12530 \\ $ 

 $=125.3 \\ $ 

Hence, the mean deviation about the mean is given by
  $M.D\left( \overline{x} \right)=\dfrac{1}{N}\sum\limits_{i=1}^{6}{{{f}_{i}}\left| {{x}_{i}}-\overline{x} \right|} \\ $ 

 $=\dfrac{1}{100}\times 1128.8 \\ $ 

 $=11.28 \\ $ 

.

11. Use the following data including ages of $\mathbf{100}$ men to find the mean deviation about the median.

Age 

Number of men

$\mathbf{16}-\mathbf{20}$

$\mathbf{5}$

$\mathbf{21}-\mathbf{25}$

$\mathbf{6}$

$\mathbf{26}-\mathbf{30}$

$\mathbf{12}$

$\mathbf{31}-\mathbf{35}$

$\mathbf{14}$

$\mathbf{36}-\mathbf{40}$

$\mathbf{26}$

$\mathbf{41}-\mathbf{45}$

$\mathbf{12}$

$\mathbf{46}-\mathbf{50}$

$\mathbf{16}$

$\mathbf{51}-\mathbf{55}$

$\mathbf{9}$


Ans.

Observe that the provided data is not continuous. So, to convert these discrete data to continuous data, subtract $0.5$ from the lower limit and add $0.5$ to the upper limit of the ages.

So, consider the following table of data.

Age

Number 

of men ${{f}_{i}}$

Cumulative frequency

(c.f.)

Mid-point

${{x}_{i}}$

$\left| {{x}_{i}}-Med. \right|$

${{f}_{i}}\left| {{x}_{i}}-Med. \right|$

$15.5-20.5$

$5$

$5$

$18$

$20$

$100$

$20.5-25.5$

$6$

$11$

$23$

$15$

$90$

$25.5-30.5$

$12$

$23$

$28$

$10$

$120$

$30.5-35.5$

$14$

$37$

$33$

$5$

$70$

$35.5-40.5$

$26$

$63$

$38$

$0$

$0$

$40.5-45.5$

$12$

$75$

$43$

$5$

$60$

$45.5-50.5$

$16$

$91$

$48$

$10$

$160$

$50.5-55.5$

$9$

$100$

$53$

$15$

$135$

Total

$100$




$735$


Therefore, here $N=100$ and it is an even number.

So, $\dfrac{N}{2}=\dfrac{100}{2}=50$. That is, the cumulative frequency greater than $50$ is $63$. Thus, the median class $=$$35.5-40.5$.

Now, we known that,

Median, $M=l+\dfrac{\dfrac{N}{2}-C}{f}\times h$

We have $l=35.5$, $C=37$, $f=26$, $h=5$. Also, $N=100$.

Therefore,

$M=35.5+\dfrac{50-37}{26}\times 5=35.5+\dfrac{13\times 5}{26}=35.5+2.5=38$.

Hence, the mean deviation about the median for the given data is

$M.D.\left( M \right)=\dfrac{1}{N}\sum\limits_{i=1}^{8}{{{f}_{i}}\left| {{x}_{i}}-M \right|} \\ $ 

 $=\dfrac{1}{100}\times 735 \\ $ 

 $=7.35 \\ $ 


NCERT Solutions for Class 11 Maths Chapters

 

NCERT Solution Class 11 Maths of Chapter 15 All Exercises

Chapter 15 - Statistics Exercises in PDF Format

Exercise 15.1

12 Questions & Solutions

Exercise 15.2

10 Questions & Solutions

Exercise 15.3

5 Questions & Solutions

Miscellaneous Exercise

7 Questions & Solutions


NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.1

Opting for the NCERT solutions for Ex 15.1 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 15.1 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 15 Exercise 15.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 15 Exercise 15.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 15 Exercise 15.1 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.


Important Topics Covered in Exercise 15.1 of Class 11 Maths NCERT Solutions

Exercise 15.1 of Class 11 Maths NCERT Solutions is based on the mean deviation which is a measure of dispersion. There are four main types of measures of dispersion which are range, standard deviation, and quartile deviation. The range of any given data gives us an overall idea of variability but does not tell us about the dispersion of the data from a measure of central tendency. For this reason, we use the standard deviation and mean. These measures can be performed on both grouped data and ungrouped data. Grouped data can be categorised into two types that are discrete frequency distribution and continuous frequency distribution. 

This exercise consists of questions on finding the mean deviation about the mean and mean deviation about the median. The solution provided in this exercise will be helpful for the students to have a better understanding of the concepts. It will be also helpful to build a solid base for learning advanced topics in Statistics.

FAQs on NCERT Solutions for Class 11 Maths Chapter 15: Statistics - Exercise 15.1

1. How many questions are there in Exercise 15.1 of Chapter 15 of Class 11 Maths, Statistics, according to NCERT Solutions?

There are a total of 11 questions in NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.1) Exercise 15.1. NCERT questions can be viewed as a helpful tool because they are filled with methods that will enable pupils to get better outcomes. To help students with their exam preparation, Vedantu provides free NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.1) Exercise 15.1.

2. Where can I obtain the PDF of the NCERT Solutions for Exercise 15.1 of Chapter 15 of Statistics in Class 11 Math?

The best online educational resource in India, Vedantu, provides students with the PDF of NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.1) Exercise 15.1. Top subject specialists developed these answers in accordance with the most recent CBSE curriculum. The solutions are presented step-by-step and are nicely structured.

3. What ideas are covered in Exercise 15.1 of NCERT Solutions for Class 11 Mathematical Statistics Chapter 15 (Ex 15.1)?

The concepts addressed in NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.1) Exercise 15.1 are measures of dispersion, mean deviation, and variance of grouped and ungrouped data. Each idea presented in this chapter ought to be clearly understood by students. Both board exams and other competitive tests, like JEE Advance and JEE Mains, will benefit from it.

4. Why does one need to study the NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.1) Exercise 15.1?

Practicing the NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.1) Exercise 15.1 becomes crucial because many of the questions in the final exams are based on it. These will help you to fully understand the concept. You'll become more efficient and there will be fewer possibilities of making errors in your calculations.

5. How do I get a free copy of the NCERT Solutions for Exercise 15.1 in Class 11 Maths Chapter 15 Statistics?

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