Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 18 Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are provided here with simple step-by-step explanations. These solutions for Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are extremely popular among Class 10 students for Maths Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Page No 859:

Mean of given observations =

Hence, the value of x is 7.

#### Page No 859:

​Mean of given observations =

Mean of 25 observations = 27

∴ Sum of 25 observations = 27 × 25 = 675

If 7 is subtracted from every number, then the sum = 675 − (25 × 7)
= 675 − 175
= 500

Then, new mean = $\frac{500}{25}=20$

Thus, the new mean will be 20.

#### Page No 859:

Here, h = 20

Let the assumed mean, A be 60.

 Class Mid-Values(xi) Frequency (fi) ${\mathbit{d}}_{\mathit{i}}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{60}$ ${\mathbit{u}}_{\mathit{i}}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{60}}{\mathbf{20}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 10−30 20 15 −40 −2 −30 30−50 40 18 −20 −1 −18 50−70 60 25 0 0 0 70−90 80 10 20 1 10 90−110 100 2 40 2 4 $\underset{}{\sum {f}_{i}=70}$ $\underset{}{\sum {f}_{i}{u}_{i}=-34}$

We know

Mean $=A+h×\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)$

∴ Mean of the given frequency distribution

Thus, the mean of the given frequency distribution is approximately 50.29.

#### Page No 859:

Here, h = 20

Let the assumed mean, A be 70.

 Class Mid-Values(xi) Frequency (fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{70}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{70}}{\mathbf{20}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 0−20 10 6 −60 −3 −18 20−40 30 8 −40 −2 −16 40−60 50 10 −20 −1 −10 60−80 70 12 0 0 0 80−100 90 6 20 1 6 100−120 110 5 40 2 10 120−140 130 3 60 3 9 $\underset{}{\sum {f}_{i}=50}$ $\underset{}{\sum {f}_{i}{u}_{i}=-19}$

We know

Mean $=A+h×\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)$

∴ Mean of the given frequency distribution

$=70+20×\left(\frac{-19}{50}\right)\phantom{\rule{0ex}{0ex}}=70-7.6\phantom{\rule{0ex}{0ex}}=62.4$

Thus, the mean of the given frequency distribution is 62.4.

#### Page No 859:

Here, h = 6

Let the assumed mean, A be 21.

 Number of days Mid-Values(xi) Number of Students (fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{21}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{21}}{\mathbf{6}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 0−6 3 10 −18 −3 −30 6−12 9 11 −12 −2 −22 12−18 15 7 −6 −1 −7 18−24 21 4 0 0 0 24−30 27 4 6 1 4 30−36 33 3 12 2 6 36−42 39 1 18 3 3 $\underset{}{\sum {f}_{i}=40}$ $\underset{}{\sum {f}_{i}{u}_{i}=-46}$

We know

Mean $=A+h×\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)$

∴ Mean number of days a student was absent

$=21+6×\left(\frac{-46}{40}\right)\phantom{\rule{0ex}{0ex}}=21-6.9\phantom{\rule{0ex}{0ex}}=14.1$

Thus, the mean number of days a student was absent is 14.1.

#### Page No 859:

Here, h = 200

Let the assumed mean, A be ₹1000.

 Expenses (in ₹) Mid-Values(xi) Number of Families(fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{1000}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{1000}}{\mathbf{200}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 500−700 600 6 −400 −2 −12 700−900 800 8 −200 −1 −8 900−1100 1000 10 0 0 0 1100−1300 1200 9 200 1 9 1300−1500 1400 7 400 2 14 $\underset{}{\sum {f}_{i}=40}$ $\underset{}{\sum {f}_{i}{u}_{i}=3}$

We know

Mean $=A+h×\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)$

∴ Mean daily expenses

$=1000+200×\left(\frac{3}{40}\right)\phantom{\rule{0ex}{0ex}}=1000+15\phantom{\rule{0ex}{0ex}}=₹1015$

Thus, the mean daily expenses is ₹1015.

#### Page No 860:

Here, h = 4

Let the assumed mean, A be 78.

 Number of Heartbeats per Minute Mid-Values(xi) Number of Patients (fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{78}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{78}}{\mathbf{4}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 64−68 66 6 −12 −3 −18 68−72 70 8 −8 −2 −16 72−76 74 10 −4 −1 −10 76−80 78 12 0 0 0 80−84 82 3 4 1 3 84−88 86 1 8 2 2 $\underset{}{\sum {f}_{i}=40}$ $\underset{}{\sum {f}_{i}{u}_{i}=-39}$

We know

Mean $=A+h×\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)$

∴ Mean heartbeats per minute for these patients

$=78+4×\left(\frac{-39}{40}\right)\phantom{\rule{0ex}{0ex}}=78-3.9\phantom{\rule{0ex}{0ex}}=74.1$

Thus, the mean heartbeats per minute for these patients is 74.1.

#### Page No 860:

Here, h = 10

Let the assumed mean, A be 35.

 Age (in years) Mid-Values(xi) Number of persons (fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{35}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{35}}{\mathbf{10}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 0−10 5 105 −30 −3 −315 10−20 15 222 −20 −2 −444 20−30 25 220 −10 −1 −220 30−40 35 138 0 0 0 40−50 45 102 10 1 102 50−60 55 113 20 2 226 60−70 65 100 30 3 300 $\underset{}{\sum {f}_{i}=1000}$ $\underset{}{\sum {f}_{i}{u}_{i}=-351}$

We know

Mean $=A+h×\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)$

∴ Mean age of the persons visiting the marketing centre on that day

$=35+10×\left(\frac{-351}{1000}\right)\phantom{\rule{0ex}{0ex}}=35-3.51\phantom{\rule{0ex}{0ex}}=31.49$

Thus, the mean age of the persons visiting the marketing centre on that day 31.49 years.

#### Page No 860:

 Class Mid-Values(xi) Frequency (fi) fi xi 0−20 10 12 120 20−40 30 15 450 40−60 50 32 1600 60−80 70 x 70x 80−100 90 13 1170 $\underset{}{\sum {f}_{i}=72+x}$ $\underset{}{\sum {f}_{i}{x}_{i}=3340+70x}$

Mean of the given frequency distribution = 53

We know

Mean $=\frac{\sum _{}{f}_{i}{x}_{i}}{\sum _{}{f}_{i}}$

$\therefore \frac{3340+70x}{72+x}=53\phantom{\rule{0ex}{0ex}}⇒3340+70x=3816+53x\phantom{\rule{0ex}{0ex}}⇒70x-53x=3816-3340$
$⇒17x=476\phantom{\rule{0ex}{0ex}}⇒x=\frac{476}{17}=28$

Thus, the value of x is 28.

#### Page No 860:

The given data is shown as follows:

 Daily pocket allowance (in ₹) Number of children (fi) Class mark (xi) fixi 11−13 7 12 84 13−15 6 14 84 15−17 9 16 144 17−19 13 18 234 19−21 f 20 20f 21−23 5 22 110 23−25 4 24 96 Total ∑ fi = 44 + f ∑ fixi = 752 + 20f

The mean of given data is given by

$\overline{x}=\frac{\sum _{i}{f}_{i}{x}_{i}}{\sum _{i}{f}_{i}}\phantom{\rule{0ex}{0ex}}⇒18=\frac{752+20f}{44+f}\phantom{\rule{0ex}{0ex}}⇒18\left(44+f\right)=752+20f\phantom{\rule{0ex}{0ex}}⇒792+18f=752+20f\phantom{\rule{0ex}{0ex}}⇒20f-18f=792-752\phantom{\rule{0ex}{0ex}}⇒2f=40\phantom{\rule{0ex}{0ex}}⇒f=20$

​Hence, the value of f is 20.

#### Page No 860:

The given data is shown as follows:

 Class Frequency (fi) Class mark (xi) fixi 0−20 7 10 70 20−40 p 30 30p 40−60 10 50 500 60−80 9 70 630 80−100 13 90 1170 Total ∑ fi = 39 + p ∑ fixi = 2370 + 30p

The mean of given data is given by

$\overline{x}=\frac{\sum _{i}{f}_{i}{x}_{i}}{\sum _{i}{f}_{i}}\phantom{\rule{0ex}{0ex}}⇒54=\frac{2370+30p}{39+p}\phantom{\rule{0ex}{0ex}}⇒54\left(39+p\right)=2370+30p\phantom{\rule{0ex}{0ex}}⇒2106+54p=2370+30p\phantom{\rule{0ex}{0ex}}⇒54p-30p=2370-2106\phantom{\rule{0ex}{0ex}}⇒24p=264\phantom{\rule{0ex}{0ex}}⇒p=11$

​Thus, the value of is 11.

#### Page No 860:

The given data is shown as follows:

 Class interval Frequency (fi) Class mark (xi) fixi 0−10 7 5 35 10−20 10 15 150 20−30 x 25 25x 30−40 13 35 455 40−50 y 45 45y 50−60 10 55 550 60−70 14 65 910 70−80 9 75 675 Total ∑ fi = 63 + x + y ∑ fixi = 2775 + 25x + 45y

Sum of the frequencies = 100

Now, The mean of given data is given by

If x = 12, then y = 37 − 12 = 25

​Thus, the value of is 12 and y is 25.

#### Page No 861:

The given data is shown as follows:

 Expenditure (in ₹) Number of families (fi) Class mark (xi) fixi 140−160 5 150 750 160−180 25 170 4250 180−200 f1 190 190f1 200−220 f2 210 210f2 220−240 5 230 1150 Total ∑ fi = 35 + f1 + f2 ∑ fixi = 6150 + 190f1 + 210f2

Sum of the frequencies = 100

Now, The mean of given data is given by

If f1 = 50, then f2 = 65 − 50 = 15

​Thus, the value of f1 is 50 and f2 is 15.

#### Page No 861:

 Class Frequency $\left({f}_{i}\right)$ Mid values $\left({x}_{i}\right)$ $\left({f}_{i}×{x}_{i}\right)$ 0-20 7 10 70 20-40 f1 30 30 f1 40-60 12 50 600 60-80 18- f1 70 1260-70 f1 80-100 8 90 720 100-120 5 110 550 $\sum {f}_{i}=50$ $\sum \left({f}_{i}×{x}_{i}\right)=3200-40{f}_{1}$

#### Page No 861:

Here, h = 80

Let the assumed mean, A be 200.

 Class Mid-Values(xi) Frequency(fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{200}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{200}}{\mathbf{80}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 0−80 40 20 −160 −2 −40 80−160 120 25 −80 −1 −25 160−240 200 x 0 0 0 240−320 280 y 80 1 y 320−400 360 10 160 2 20 $\underset{}{\sum {f}_{i}=x+y+55}$ $\underset{}{\sum {f}_{i}{u}_{i}=-45+y}$

$\underset{}{\sum {f}_{i}=x+y+55}=100$         (Given)

We know

Mean $=A+h×\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)$

$\therefore 200+80×\left(\frac{-45+y}{100}\right)=188$          (Given)

$⇒\frac{-45+y}{100}=\frac{188-200}{80}=-0.15\phantom{\rule{0ex}{0ex}}⇒-45+y=-15\phantom{\rule{0ex}{0ex}}⇒y=45-15=30$

Substituting the value of y in (1), we get

x + 30 = 45

x = 45 − 30 = 15

Thus, the missing frequencies x and y are 15 and 30, respectively.

#### Page No 861:

Using Direct method, the given data is shown as follows:

 Literacy rate (%) Number of cities (fi) Class mark (xi) fixi 45−55 4 50 200 55−65 11 60 660 65−75 12 70 840 75−85 9 80 720 85−95 4 90 360 Total ∑ fi = 40 ∑ fixi = 2780

The mean of given data is given by

​Thus, the mean literacy rate is 69.5%.

#### Page No 861:

 Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ Deviation $\left({d}_{i}\right)$ ${d}_{i}=\left({x}_{i}-25\right)$ $\left({f}_{i}×{d}_{i}\right)$ 0-10 12 5 -20 -240 10-20 18 15 -10 -180 20-30 27 25=A 0 0 30-40 20 35 10 200 40-50 17 45 20 340 50-60 6 55 30 180 $\sum {f}_{i}=100$ $\sum \left({f}_{i}×{d}_{i}\right)=300$

#### Page No 861:

 Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ Deviation $\left({d}_{i}\right)$ ${d}_{i}=\left({x}_{i}-150\right)$ $\left({f}_{i}×{d}_{i}\right)$ 100-120 10 110 -40 -400 120-140 20 130 -20 -400 140-160 30 150=A 0 0 160-180 15 170 20 300 180-200 5 190 40 200 $\sum {f}_{i}=80$ $\sum \left({f}_{i}×{d}_{i}\right)=-\text{3}00$

#### Page No 861:

 Class Frequency $\left({f}_{i}\right)$ Mid Values$\left({x}_{i}\right)$ Deviation $\left({d}_{i}\right)$ ${d}_{i}=\left({x}_{i}-50\right)$ $\left({f}_{i}×{d}_{i}\right)$ 0-20 20 10 -40 -800 20-40 35 30 -20 -700 40-60 52 50=A 0 0 60-80 44 70 20 880 80-100 38 90 40 1520 100-120 31 110 60 1860 $\sum {f}_{i}=220$ $\sum \left({f}_{i}×{d}_{i}\right)=2760$

#### Page No 862:

Let us choose a = 25, h = 10, then di = xi − 25 and ui$\frac{{x}_{i}-25}{10}$.

Using Step-deviation method, the given data is shown as follows:

 Class Frequency (fi) Class mark (xi) di = xi − 25 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{25}}{\mathbf{10}}$ fiui 0−10 7 5 −20 −2 −14 10−20 10 15 −10 −1 −10 20−30 15 25 0 0 0 30−40 8 35 10 1 8 40−50 10 45 20 2 20 Total ∑ fi = 50 ∑ fiui = 4

The mean of given data is given by

​Thus, the mean is 25.8.

#### Page No 862:

Let us choose a = 40, h = 10, then di = xi − 40 and ui = $\frac{{x}_{i}-40}{10}$.

Using Step-deviation method, the given data is shown as follows:

 Class Frequency (fi) Class mark (xi) di = xi − 40 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{40}}{\mathbf{10}}$ fiui 5−15 6 10 −30 −3 −18 15−25 10 20 −20 −2 −20 25−35 16 30 −10 −1 −16 35−45 15 40 0 0 0 45−55 24 50 10 1 24 55−65 8 60 20 2 16 65−75 7 70 30 3 21 Total ∑ fi = 86 ∑ fiui = 7

The mean of given data is given by

​Thus, the mean is 40.81.

#### Page No 862:

​Let us choose a = 202.5, h = 1, then di = xi − 202.5 and ui = $\frac{{x}_{i}-202.5}{1}$.

Using Step-deviation method, the given data is shown as follows:

 Weight (in grams) Number of packets (fi) Class mark (xi) di = xi − 202.5 ui = $\frac{{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{202}\mathbf{.}\mathbf{5}}{\mathbf{1}}$ fiui 200−201 13 200.5 −2 −2 −26 201−202 27 201.5 −1 −1 −27 202−203 18 202.5 0 0 0 203−204 10 203.5 1 1 10 204−205 1 204.5 2 2 2 205−206 1 205.5 3 3 3 Total ∑ fi = 70 ∑ fiui = −38

The mean of given data is given by

​Hence, the mean is 201.96 g.

#### Page No 862:

Let us choose a = 45, h = 10, then di = xi − 45 and ui = $\frac{{x}_{i}-45}{10}$.

Using Step-deviation method, the given data is shown as follows:

 Class Frequency (fi) Class mark (xi) di = xi − 45 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{45}}{\mathbf{10}}$ fiui 20−30 25 25 −20 −2 −50 30−40 40 35 −10 −1 −40 40−50 42 45 0 0 0 50−60 33 55 10 1 33 60−70 10 65 20 2 20 Total ∑ fi = 150 ∑ fiui = −37

The mean of given data is given by

​Thus, the mean is 42.534.

#### Page No 862:

 Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-33\right)}{6}$ $\left({f}_{i}×{u}_{i}\right)$ 18-24 6 21 −2 −12 24-30 8 27 −1 −8 30-36 12 33 = A 0 0 36-42 8 39 1 8 42-48 4 45 2 8 48-54 2 51 3 6 $\sum {f}_{i}=40$ $\sum \left({f}_{i}×{u}_{i}\right)=2$

#### Page No 862:

 Class Frequency$\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-550\right)}{20}$ $\left({f}_{i}×{u}_{i}\right)$ 500-520 14 510 −2 −28 520-540 9 530 −1 −9 540-560 5 550 = A 0 0 560-580 4 570 1 4 580-600 3 590 2 6 600-620 5 610 3 15 $\sum {f}_{i}=40$ $\sum \left({f}_{i}×{u}_{i}\right)=-12$

#### Page No 862:

Converting the series into exclusive form, we get:

 Class Frequency$\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-42\right)}{5}$ $\left({f}_{i}×{u}_{i}\right)$ 24.5-29.5 4 27 −3 −12 29.5-34.5 14 32 −2 −28 34.5-39.5 22 37 −1 −22 39.5-44.5 16 42 = A 0 0 44.5-49.5 6 47 1 6 49.5-54.5 5 52 2 10 54.5-59.5 3 57 3 9 $\sum {f}_{i}=70$ $\sum \left({f}_{i}×{u}_{i}\right)=-37$

#### Page No 863:

Converting the series into exclusive form, we get:

 Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-29.5\right)}{10}$ $\left({f}_{i}×{u}_{i}\right)$ 4.5-14.5 6 9.5 −2 −12 14.5-24.5 11 19.5 −1 −11 24.5-34.5 21 29.5 = A 0 0 34.5-44.5 23 39.5 1 23 44.5-54.5 14 49.5 2 28 54.5-64.5 5 59.5 3 15 $\sum {f}_{i}=80$ $\sum \left({f}_{i}×{u}_{i}\right)=43$

#### Page No 863:

Let us choose a = 92, h = 5, then di = xi − 92 and ui = $\frac{{x}_{i}-92}{5}$.

Using Step-deviation method, the given data is shown as follows:

 Weight (in grams) Number of eggs (fi) Class mark (xi) di = xi − 92 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{92}}{\mathbf{5}}$ fiui 74.5−79.5 4 77 −15 −3 −12 79.5−84.5 9 82 −10 −2 −18 84.5−89.5 13 87 −5 −1 −13 89.5−94.5 17 92 0 0 0 94.5−99.5 12 97 5 1 12 99.5−104.5 3 102 10 2 6 104.5−109.5 2 107 15 3 6 Total ∑ fi = 60 ∑ fiui = −19

The mean of given data is given by

​Thus, the mean weight to the nearest gram is 90 g.

#### Page No 863:

Let us choose a = 17.5, h = 5, then di = xi − 17.5 and ui = $\frac{{x}_{i}-17.5}{5}$.

Using Step-deviation method, the given data is shown as follows:

 Marks Number of students (cf) Frequency (fi) Class mark (xi) di = xi − 17.5 ui = $\frac{{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{17}\mathbf{.}\mathbf{5}}{\mathbf{5}}$ fiui 0−5 3 3 2.5 −15 −3 −9 5−10 10 7 7.5 −10 −2 −14 10−15 25 15 12.5 −5 −1 −15 15−20 49 24 17.5 0 0 0 20−25 65 16 22.5 5 1 16 25−30 73 8 27.5 10 2 16 30−35 78 5 32.5 15 3 15 35−40 80 2 37.5 20 4 8 Total ∑ fi = 80 ∑ fiui = 17

The mean of given data is given by

​Thus, the mean marks correct to 2 decimal places is 18.56.

#### Page No 870:

We prepare the cumulative frequency table, as shown below:

 Age (in years) Number of patients (fi) Cumulative Frequency (cf) 0−15 5 5 15−30 20 25 30−45 40 65 45−60 50 115 60−75 25 140 Total N = ∑ fi = 140

Now, N = 140 $⇒\frac{N}{2}=70$.

The cumulative frequency just greater than 70 is 115 and the corresponding class is 45−60.

Thus, the median class is 45−60.

∴ l = 45, h = 15, f = 50, N = 140 and cf = 65.

Now,

Hence, the median age is 46.5 years.

#### Page No 870:

 Class Frequency (f) Cumulative frequency 0-7 3 3 7-14 4 7 14-21 7 14 21-28 11 25 28-35 0 25 35-42 16 41 42-49 9 50 N=∑f=50

#### Page No 870:

 Class Frequency(f) Cumulative frequency 0-100 40 40 100-200 32 72 200-300 48 120 300-400 22 142 400-500 8 150 N=∑f=150

#### Page No 870:

 Class Frequency(f) Cumulative frequency 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 N=∑f=49

#### Page No 870:

 Class Frequency(f) Cumulative frequency 65-85 4 4 85-105 5 9 105-125 13 22 125-145 20 42 145-165 14 56 165-185 7 63 185-205 4 67 N=∑f=67

#### Page No 870:

 Class Frequency(f) Cumulative frequency 135-140 6 6 140=145 10 16 145-150 18 34 150-155 22 56 155-160 20 76 160-165 15 91 165-170 6 97 170-175 3 100 N=∑f=100

#### Page No 870:

 Class Frequency (fi) c.f 0-10 5 5 10-20 25 30 20-30 x x+30 30-40 18 x+48 40-50 7 x+55

#### Page No 871:

We prepare the cumulative frequency table, as shown below:

 Class Frequency (fi) Cumulative frequency (cf) 0−5 12 12 5−10 a 12 + a 10−15 12 24 + a 15−20 15 39 + a 20−25 b 39 + a + b 25−30 6 45 + a + b 30−35 6 51 + a + b 35−40 4 55 + a + b Total N = ∑fi = 70

Let a and b be the missing frequencies of class intervals 5−10 and 20−25 respectively. Then,

55 a + = 70 ⇒ a + = 15     ...(1)

Median is 16, which lies in 15−20. So, the median class is 15−20.

l = 15, h = 5, N = 70, f = 15 and cf = 24 + a

Now,

$\mathrm{Median}=l+\left(\frac{\frac{N}{2}-cf}{f}\right)×h\phantom{\rule{0ex}{0ex}}⇒16=15+\left(\frac{\frac{70}{2}-\left(24+a\right)}{15}\right)×5\phantom{\rule{0ex}{0ex}}⇒16=15+\left(\frac{35-24-a}{3}\right)\phantom{\rule{0ex}{0ex}}⇒16=15+\left(\frac{11-a}{3}\right)\phantom{\rule{0ex}{0ex}}⇒16-15=\frac{11-a}{3}\phantom{\rule{0ex}{0ex}}⇒1×3=11-a\phantom{\rule{0ex}{0ex}}⇒a=11-3\phantom{\rule{0ex}{0ex}}⇒a=8$

b = 15 a    [From (1)]
b = 15 − 8
b = 7

Hence, a = 8 and b = 7.

#### Page No 871:

We prepare the cumulative frequency table, as shown below:

 Runs scored Number of batsmen (fi) Cumulative frequency (cf) 2500−3500 5 5 3500−4500 x 5 + x 4500−5500 y 5 + x + y 5500−6500 12 17 + x + y 6500−7500 6 23 + x + y 7500−8500 2 25 + x + y Total N = ∑fi = 60

Let x and y be the missing frequencies of class intervals 3500−4500 and 4500−5500 respectively. Then,

25 x + = 60 ⇒ x + = 35     ...(1)

Median is 5000, which lies in 4500−5500. So, the median class is 4500−5500.

∴ l = 4500, h = 1000, N = 60, f = y and cf = 5 + x

Now,

∴ y = 35 − x    [From (1)]
⇒ y = 35 − 15
⇒ y = 20

Hence, x = 15 and y = 20.

#### Page No 871:

 Class Frequency(f) Cumulative frequency 0-10 ${f}_{1}$ ${f}_{1}$ 10-20 5 ${f}_{1}$+5 20-30 9 ${f}_{1}$+14 30-40 12 ${f}_{1}$+26 40-50 ${f}_{2}$ ${f}_{1}+{f}_{2}$+26 50-60 3 ${f}_{1}+{f}_{2}$+29 60-70 2 ${f}_{1}+{f}_{2}$+31 N=∑f=40

#### Page No 871:

First, we will convert the data into exclusive form.

 Class Frequency(f) Cumulative frequency 18.5-25.5 35 35 25.5-32.5 96 131 32.5-39.5 68 199 39.5-46.5 102 301 46.5-53.5 35 336 53.5-60.5 4 340 N=∑f=340

#### Page No 871:

Converting the given data into exclusive form, we get:

 Class Frequency(f) Cumulative frequency 60.5-70.5 5 5 70.5-80.5 15 20 80.5-90.5 20 40 90.5-100.5 30 70 100.5-110.5 20 90 110.5-120.5 8 98 N=∑f=98

#### Page No 871:

Converting into exclusive form, we get:

 Class Frequency(f) Cumulative frequency 0.5-5.5 7 7 5.5-10.5 10 17 10.5-15.5 16 33 15.5-20.5 32 65 20.5-25.5 24 89 25.5-30.5 16 105 30.5-35.5 11 116 35.5-40.5 5 121 40.5-45.5 2 123 N=∑f=123

#### Page No 871:

 Class Cumulative frequency Frequency (f) 0-10 12 12 10-20 32 20 20-30 57 25 30-40 80 23 40-50 92 12 50-60 116 24 60-70 164 48 70-80 200 36 N=∑f=200

#### Page No 877:

Here the maximum class frequency is 45, and the class corresponding to this frequency is 30−40. So, the modal class is 30−40.

Now,

modal class = 30−40, lower limit (l) of modal class = 30, class size (h) = 10,

frequency (f1) of the modal class = 45,

frequency (f0) of class preceding the modal class = 35,

frequency (
f2) of class succeeding the modal class = 25

Now, let us substitute these values in the formula:

Hence, the mode is 33.33.

#### Page No 877:

Here the maximum class frequency is 28, and the class corresponding to this frequency is 40−60. So, the modal class is 40−60.

Now,

Modal class = 40−60, lower limit (l) of modal class = 40, class size (h) = 20,

frequency (f1) of the modal class = 28,

frequency (f0) of class preceding the modal class = 16,

frequency (f2
) of class succeeding the modal class = 20.

Now, let us substitute these values in the formula:

Hence, the mode is 52.

#### Page No 878:

Here the maximum class frequency is 20, and the class corresponding to this frequency is 160−165. So, the modal class is 160−165.

Now,

Modal class =
160−165, lower limit (l) of modal class = 160, class size (h) = 5,

frequency (f1) of the modal class = 20,

frequency (f0) of class preceding the modal class = 8,

frequency (f2) of class succeeding the modal class = 12.

Now, let us substitute these values in the formula:

Hence, the mode is 163.

It represents that the height of maximum number of students is 163 cm.

Now, to find the mean let us put the data in the table given below:

 Height (in cm) Number of students (fi) Class mark (xi) fixi 150−155 15 152.5 2287.5 155−160 8 157.5 1260 160−165 20 162.5 3250 165−170 12 167.5 2010 170−175 5 172.5 862.5 Total ∑fi = 60 ∑fixi =  9670

Thus, mean of the given data is 161.17.

It represents that on an average, the height of a student is 161.17 cm.

#### Page No 878:

As the class 26-30 has the maximum frequency, it is the modal class.

#### Page No 878:

As the class 1500-2000 has the maximum frequency, it is the modal class.

Hence, mode = Rs 1820

#### Page No 878:

As the class 5000-10000 has the maximum frequency, it is the modal class.

Hence, mode = Rs 7727.27

#### Page No 879:

As the class 15-20 has the maximum frequency, it is the modal class.

Hence, mode=17.3 years

#### Page No 879:

As the class 85-95 has the maximum frequency, it is the modal class.

Hence, mode=85.71

#### Page No 879:

Clearly, we have to find the mode of the data. The given data is an inclusive series. So, we will convert it to an exclusive form as given below:

 Class interval 0.5-5.5 5.5-10.5 10.5-15.5 15.5-20.5 20.5-25.5 25.5-30.5 30.5-35.5 35.5-40.5 40.5-45.5 45.5-50.5 Frequency 3 8 13 18 28 20 13 8 6 4

As the class 20.5-25.5 has the maximum frequency, it is the modal class.

Hence, mode=23.28

#### Page No 879:

It is given that the sum of frequencies is 181.

x + 15 + 18 + 30 + 50 + 48 + x = 181
⇒ 2x + 161 = 181
⇒ 2x = 181 − 161
⇒ 2x = 20
x = 10

Thus, x = 10.

Here the maximum class frequency is 50, and the class corresponding to this frequency is 13−15. So, the modal class is 13−15.

Now,

Modal class =
13−15, lower limit (l) of modal class = 13, class size (h) = 2,

frequency (f1) of the modal class = 50,

frequency (f0) of class preceding the modal class = 30,

frequency (f2) of class succeeding the modal class = 48.

Now, let us substitute these values in the formula:

Hence, the mode is 14.82.

#### Page No 881:

To find the mean let us put the data in the table given below:

 Class Frequency (fi) Class mark (xi) fixi 0−10 4 5 20 10−20 4 15 60 20−30 7 25 175 30−40 10 35 350 40−50 12 45 540 50−60 8 55 440 60−70 5 65 325 Total ∑fi = 50 ∑fixi =  1910

Thus, mean of the given data is 38.2.

Now, to find the median let us put the data in the table given below:

 Class Frequency (fi) Cumulative frequency (cf) 0−10 4 4 10−20 4 8 20−30 7 15 30−40 10 25 40−50 12 37 50−60 8 45 60−70 5 50 Total N = ∑fi = 50

Now, N = 50 $⇒\frac{N}{2}=25$.

The cumulative frequency just greater than 25 is 37, and the corresponding class is 40−50.

Thus, the median class is 40−50.

l = 40, h = 10, N = 50, f = 12 and cf = 25.

Now,

Thus, the median is 40.

We know that,
Mode = 3(median) − 2(mean)
= 3 × 40 − 2 × 38.2
= 120 − 76.4
= 43.6

Hence, Mean = 38.2, Median = 40 and Mode = 43.6

#### Page No 881:

To find the mean let us put the data in the table given below:

 Class Frequency (fi) Class mark (xi) fixi 0−20 6 10 60 20−40 8 30 240 40−60 10 50 500 60−80 12 70 840 80−100 6 90 540 100−120 5 110 550 120−140 3 130 390 Total ∑fi = 50 ∑fixi =  3120

Thus, mean of the given data is 62.4.

Now, to find the median let us put the data in the table given below:

 Class Frequency (fi) Cumulative frequency (cf) 0−20 6 6 20−40 8 14 40−60 10 24 60−80 12 36 80−100 6 42 100−120 5 47 120−140 3 50 Total N = ∑fi = 50

Now, N = 50 $⇒\frac{N}{2}=25$.

The cumulative frequency just greater than 25 is 36, and the corresponding class is 60−80.

Thus, the median class is 60−80.

∴ l = 60, h = 20, N = 50, f = 12 and cf = 24.

Now,

Thus, the median is 61.67.

We know that,
Mode = 3(median) − 2(mean)
= 3 × 61.67 − 2 × 62.4
= 185.01 − 124.8
= 60.21

Hence, Mean = 62.4, Median = 61.67 and Mode = 60.21.

#### Page No 882:

To find the mean let us put the data in the table given below:

 Class Frequency (fi) Class mark (xi) fixi 0−50 2 25 50 50−100 3 75 225 100−150 5 125 625 150−200 6 175 1050 200−250 5 225 1125 250−300 3 275 825 300−350 1 325 325 Total ∑fi = 25 ∑fixi =  4225

Thus, mean of the given data is 169.

Now, to find the median let us put the data in the table given below:

 Class Frequency (fi) Cumulative frequency (cf) 0−50 2 2 50−100 3 5 100−150 5 10 150−200 6 16 200−250 5 21 250−300 3 24 300−350 1 25 Total N = ∑fi = 25

Now, N = 25 $⇒\frac{N}{2}=12.5$.

The cumulative frequency just greater than 12.5 is 16, and the corresponding class is 150−200.

Thus, the median class is 150−200.

∴ l = 150, h = 50, N = 25, f = 6 and cf = 10.

Now,

Thus, the median is 170.83.

We know that,
Mode = 3(median) − 2(mean)
= 3 × 170.83 − 2 × 169
= 512.49 − 338
= 174.49

Hence, Mean = 169, Median = 170.83 and Mode = 174.49

#### Page No 882:

To find the mean let us put the data in the table given below:

 Marks obtained Number of students (fi) Class mark (xi) fixi 25−35 7 30 210 35−45 31 40 1240 45−55 33 50 1650 55−65 17 60 1020 65−75 11 70 770 75−85 1 80 80 Total ∑fi = 100 ∑fixi =  4970

Thus, mean of the given data is 49.7.

Now, to find the median let us put the data in the table given below:

 Class Frequency (fi) Cumulative frequency (cf) 25−35 7 7 35−45 31 38 45−55 33 71 55−65 17 88 65−75 11 99 75−85 1 100 Total N = ∑fi = 100

Now, N = 100 $⇒\frac{N}{2}=50$.

The cumulative frequency just greater than 50 is 71, and the corresponding class is 45−55.

Thus, the median class is 45−55.

∴ l = 45, h = 10, N = 100, f = 33 and cf = 38.

Now,

Thus, the median is 48.64.

We know that,
Mode = 3(median) − 2(mean)
= 3 × 48.64 − 2 × 49.70
= 145.92 − 99.4
= 46.52

Hence, Mean = 49.70, Median = 48.64 and Mode = 46.52.

#### Page No 882:

We have the following:

 Height in cm Mid value $\left({x}_{i}\right)$ Frequency $\left({f}_{i}\right)$ Cumulative frequency ${f}_{i}×{x}_{i}$ 120-130 125 2 2 250 130-140 135 8 10 1080 140-150 145 12 22 1740 150-160 155 20 42 3100 160-170 165 8 50 1320 ∑${f}_{i}=50$ ∑$\left({f}_{i}×{x}_{i}\right)=$7490
Mean, $\overline{x}$=$\frac{\sum \left({f}_{i}×{x}_{i}\right)}{\sum {f}_{i}}$
=$\frac{7490}{50}$

=149.8

The cumulative frequency just greater than 25 is 42 and the corresponding class is 150-160.
Thus, the median class is 150-160.
of preceding class = 22 and $\frac{N}{2}=25$

=151.5
∴ Mode = 3(Median) (Mean)
= 3$×151.5-2×149.8$
= 154.9

#### Page No 882:

We have the following:

 Daily income Mid value $\left({x}_{i}\right)$ Frequency $\left({f}_{i}\right)$ Cumulative frequency ${f}_{i}×{x}_{i}$ 100-120 110 12 12 1320 120-140 130 14 26 1820 140-160 150 8 34 1200 160-180 170 6 40 1020 180-200 190 10 50 1900 ∑${f}_{i}=50$ ∑$\left({f}_{i}×{x}_{i}\right)=$7260

Mean, $\overline{x}$=$\frac{\sum \left({f}_{i}×{x}_{i}\right)}{\sum {f}_{i}}$
=$\frac{7260}{50}$

=145.2

The cumulative frequency just greater than 25 is 26 and the corresponding class is 120-140.
Thus, the median class is 120-140.
of preceding class = 12 and$\frac{N}{2}=25$
Median, ${M}_{e}=l+\left\{h×\frac{\left(\frac{N}{2}-c\right)}{f}\right\}$
$=120+\left\{20×\frac{\left(25-12\right)}{14}\right\}\phantom{\rule{0ex}{0ex}}=\left(120+20×\frac{13}{14}\right)$

=138.57
Mode = 3(Median) $-$ 2(mean)
$=3×138.57-2×145.2\phantom{\rule{0ex}{0ex}}=125.31$

#### Page No 882:

We have the following:

 Daily income Mid value Frequency$\left({f}_{i}\right)$ Cumulative frequency ${f}_{i}×{x}_{i}$ 100-150 125 6 6 750 150-200 175 7 13 1225 200-250 225 12 25 2700 250-300 275 3 28 825 300-350 325 2 30 650 ∑${f}_{i}=30$ ∑$\left({f}_{i}×{x}_{i}\right)=$6150

Mean, $\overline{x}$=$\frac{\sum \left({f}_{i}×{x}_{i}\right)}{\sum {f}_{i}}$
=$\frac{6150}{30}$

=205

The cumulative frequency just greater than 15 is 25 and the corresponding class is 200-250.
Thus, the median class is 200-250.
of preceding class = 13 and $\frac{N}{2}=15$
Median, ${M}_{e}=l+\left\{h×\frac{\left(\frac{N}{2}-c\right)}{f}\right\}$
$=200+\left\{50×\frac{\left(15-13\right)}{12}\right\}\phantom{\rule{0ex}{0ex}}=\left(200+50×\frac{2}{12}\right)$
=200 + 8.33
=208.33

#### Page No 892:

The frequency distribution table of less than type is given as follows:

 Marks (upper class limits) Cumulative frequency (cf) Less than 10 5 Less than 20 5 + 3 = 8 Less than 30 8 + 4 = 12 Less than 40 12 + 3 = 15 Less than 50 15 + 3 = 18 Less than 60 18 + 4 = 22 Less than 70 22 + 7 = 29 Less than 80 29 + 9 = 38 Less than 90 38 + 7 = 45 Less than 100 45 + 8 = 53

Taking upper class limits of class intervals on x−axis and their respective frequencies on y−axis, its ogive can be drawn as follows: Here, N = 53 $⇒\frac{N}{2}=26.5$.

Mark the point A whose ordinate is 26.5 and its x−coordinate is 66.4. Thus, median of the data is 66.4.

#### Page No 892:

Taking upper class limits of class intervals on x−axis and their respective frequencies on y−axis, its ogive can be drawn as follows: Here, N = 80 $⇒\frac{N}{2}=40$.

Mark the point A whose ordinate is 40 and its x−coordinate is 76. Thus, median of the data is 76.

#### Page No 893:

The frequency distribution table of more than type is as follows:

 Marks (lower class limits) Cumulative frequency (cf) More  than 0 96 + 4 = 100 More  than 10 90 + 6 = 96 More  than 20 80 + 10 = 90 More  than 30 70 + 10 = 80 More  than 40 45 + 25 = 70 More  than 50 23 + 22 = 45 More  than 60 18 + 5 = 23 More  than 70 5

Taking the lower class limits on x−axis and their respective cumulative frequencies on y−axis, its ogive can be obtained as follows: #### Page No 893:

The frequency distribution table of more than type is as follows:

 Height (in cm) (lower class limits) Cumulative frequency (cf) More  than 135 5 + 45 = 50 More  than 140 8 + 37 = 45 More  than 145 9 + 28 = 37 More  than 150 12 + 16 = 28 More  than 155 14 + 2 = 16 More  than 160 2

Taking the lower class limits on x−axis and their respective cumulative frequencies on y−axis, its ogive can be obtained as follows: #### Page No 893:

The frequency distribution table of more than type is as follows:

 Height (in cm) (lower class limits) Cumulative frequency (cf) More  than 140 3 + 153 = 156 More  than 160 8 + 145 = 153 More  than 180 15 + 130 = 145 More  than 200 40 + 90 = 130 More  than 220 50 + 40 = 90 More  than 240 30 + 10 = 40 More  than 260 10

Taking the lower class limits on x−axis and their respective cumulative frequencies on y−axis, its ogive can be obtained as follows: #### Page No 893: Here, N = 100 $⇒\frac{N}{2}=50.$