NCERT Solutions for Class 10 Maths Chapter 14 - Statistics
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Exercise 14.1
1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
| Number of plants | Number of housesn |
| 0 – 2 | 1 |
| 2 – 4 | 2 |
| 4 – 6 | 1 |
| 6 – 8 | 5 |
| 8 – 10 | 6 |
| 10 – 12 | 2 |
| 12 – 14 | 3 |
Which method did you use for finding the mean and why?
Sol. Let us find mean of the data by direct method because the figures are small.
| Class (Number of plants) | Frequency fi (Number of houses) | Class marks xi | fi × xi |
| 0–2 | 1 | 1 | 1 |
| 2-4 | 2 | 3 | 6 |
| 4-6 | 1 | 5 | 5 |
| 6-8 | 5 | 7 | 35 |
| 8-10 | 6 | 9 | 54 |
| 10-12 | 2 | 11 | 22 |
| 12-14 | 3 | 13 | 39 |
| Total | N=20 | 162 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
$$\text{We have, N =}\space\sum_{i=1}^{7}f_i=20\\\text{and\space}\sum_{i=1}^{7}f_ix_i=162$$
Then, mean of the data of
$$\bar x=\frac{1}{\text{N}}×\sum_{i=1}^{7}f_ix_i\\=\frac{1}{20}×162=8.1$$
Hence, the required mean of the data is 8.1 plants.
2. Consider the following distribution of daily wages of 50 workers of a factory.
| Daily wages (in ₹) | Number of workers |
| 100 – 120 | 12 |
| 120 – 140 | 14 |
| 140 – 160 | 8 |
| 160 – 180 | 6 |
| 160 – 180 | 6 |
| 180 – 200 | 10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
Sol. Here, we use step deviation method for finding the mean of the data.
| Daily wages (in ₹) | Number of workers fi | Class mark xi | di = xi – 150 | $$u_i=\frac{d_i}{h}=\frac{x_i-150}{20}$$ | fi ui |
| 100-120 | 12 | 110 | – 40 | – 2 | – 24 |
| 120-140 | 14 | 130 | -20 | -1 | -14 |
| 140-160 | 8 | 150=a | 0 | 0 | 0 |
| 160-180 | 6 | 170 | 20 | 1 | 6 |
| 180-200 | 10 | 190 | 40 | 2 | 20 |
| Total | N = 50 | – 12 |
Here, a (Assumed mean) = 150
h (Class width) = 20
N = 50
and Σfiui = – 12
By step deviation method, we have
$$\text{Mean =}a+h×\frac{1}{h}\sum f_iu_i\\=\begin{Bmatrix}150+20×\frac{1}{50}×(-12)\end{Bmatrix}\\\begin{Bmatrix}150-\frac{24}{5}\end{Bmatrix}$$
= {150 – 4.80} = ₹ 145.20
3. The following distribution show the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.
| Daily pocket allowance (in ₹) | Number of children |
| 11 – 13 | 7 |
| 13 – 15 | 6 |
| 15 – 17 | 9 |
| 17 – 19 | 13 |
| 19 – 21 | f |
| 21 – 23 | 5 |
| 23 – 25 | 4 |
Sol.
| Daily pocket allowance (in ₹) | Number of children fi | Class mark xi | di = xi – 18 | fi×di |
| 11 – 13 | 7 | 12 | -6 | -42 |
| 13 – 15 | 6 | 14 | -4 | -24 |
| 15 – 17 | 9 | 16 | -2 | -18 |
| 17 – 19 | 13 | 18=a | 0 | 0 |
| 19 – 21 | f | 20 | 2 | 2f |
| 21 – 23 | 5 | 22 | 4 | 20 |
| 23 – 25 | 4 | 24 | 6 | 24 |
| Total | N = 44 + f | 2f – 40 |
Here,
$$\bar x=₹\space18$$
From the table, we have
a (assumed mean) = 18, N = 44 + f
$$\text{Now,\space mean}=a+\frac{1}{\text{N}}×\Sigma f_id_i$$
Then, substituting the value as given above, we have
$$18=18+\frac{1}{44-f}×(2f-40)\\\Rarr\space 0=\frac{2f-40}{44+f}$$
⇒ 2f = 40
⇒ f = 20
4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
| Number of heart beats per minute | Number of women |
| 65 – 68 | 2 |
| 68 – 71 | 4 |
| 71 – 74 | 3 |
| 74 – 77 | 8 |
| 74 – 77 | 8 |
| 77 – 80 | 7 |
| 80 – 83 | 4 |
| 83 – 86 | 2 |
Sol.
| Number of heart beats per minute | Number of women fi | Class mark xi | $$u_i=\frac{x_i-a}{h}=\frac{x_i-75.5}{3}$$ | fi ui |
| 65 – 68 | 2 | 66.5 | $$u_i=\frac{66.5-75.5}{3}=-3$$ | – 6 |
| 68 – 71 | 4 | 69.5 | $$u_2=\frac{69.5-75.5}{3}=-2$$ | -8 |
| 71 – 74 | 3 | 72.5 | $$u_3=\frac{72.5-75.5}{3}=-3$$ | -3 |
| 74 – 77 | 8 | 75.5 = a | $$u_4=\frac{75.5-75.5}{3}=0$$ | 0 |
| 77 – 80 | 7 | 78.5 | $$u_5=\frac{78.5-75.5}{3}=1$$ | 7 |
| 80-83 | 4 | 81.5 | $$u_6=\frac{81.5-75.5}{3}=2$$ | 8 |
| 83 – 86 | 2 | 84.5 | $$u_7=\frac{84.5-75.5}{3}=3$$ | 6 |
| Total | N = 30 | 4 |
Here, a (Assumed mean) = 75.5
h (Class width) = 3
N = 30
By step deviation method,
$$\text{Mean}=a+h×\frac{1}{h}\Sigma f_iu_i\\=75.5+3×\frac{1}{30}×4$$
= 75.5 + 0.4 = 75.9 heart beats
5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
| Number of mangoes | Number of boxes |
| 50 – 52 | 15 |
| 53 – 55 | 110 |
| 56 – 58 | 135 |
| 59 – 61 | 115 |
| 62 – 64 | 25 |
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Sol. Since, given data is not continuous, so we subtract and add 0.5 respectively in the lower and upper limit of each class.
| Number of mangoes | Number of boxes (fi) | Class mark (xi) | $$u_i=\frac{x_i-a}{h}=\frac{x_i-57}{3}$$ | fi ui |
| 49.5 – 52.5 | 15 | 51 | -2 | -30 |
| 52.5 – 55.5 | 110 | 54 | -1 | -110 |
| 55.5 – 58.5 | 135 | a = 57 | 0 | 0 |
| 58.5 – 61.5 | 115 | 60 | 1 | 115 |
| 61.5 – 64.5 | 25 | 63 | 2 | 50 |
| Total | N = 400 | 25 |
Here, a = 57
h = 3
N = 400
$$\text{Now,\space \text{Mean}}=a+h×\frac{1}{h}×\Sigma f_iu_i\\=57+3×\frac{1}{400}×25\\=57+\frac{75}{400}$$
= 57 + 0.1875 = 57 + 0.19 = 57.19
6. The table below shows the daily expenditure of food of 25 households in a locality.
| Daily expenditure (in ₹) | Number of households |
| 100 – 150 | 4 |
| 150 – 200 | 5 |
| 200 – 250 | 12 |
| 250 – 300 | 2 |
| 300 – 350 | 2 |
Find the mean daily expenditure of food by a suitable method.
Sol.
| Daily expenditure (in ₹) | Number of households (fi) | Class mark (xi) | $$u_i=\frac{x_i-a}{h}=\frac{x_i-225}{50}$$ | fi ui |
| 100 – 150 | 4 | 125 | -2 | -8 |
| 150 – 200 | 5 | 175 | -1 | -5 |
| 200 – 250 | 12 | a = 225 | 0 | 0 |
| 250 – 300 | 2 | 275 | 1 | 2 |
| 300 – 350 | 2 | 325 | 2 | 4 |
| TOTAL | N = 25 | – 7 |
Here,
a = 225
h = 50
N = 25
$$\text{Now,\space Mean =a+h}\space×\frac{1}{h}×\Sigma f_iu_i\\=225+50×\frac{1}{25}×(\normalsize-7)$$
= 225 – 14 = ₹ 211
7. To find out the concentration of SO2 in the air (in parts per million i.e., ppm), the data was collected for 30 localities in a certain city and is presented below.
| Concentration of SO2 (in ppm) | Frequency |
| 0.00 – 0.04 | 4 |
| 0.04 – 0.08 | 9 |
| 0.08 – 0.12 | 9 |
| 0.12 – 0.16 | 2 |
| 0.16 – 0.20 | 4 |
| 0.20 – 0.24 | 2 |
Find the mean concentration of SO2 in the air.
Sol.
| Concentration of SO2 (in ppm) | fi | Class mark (xi) | $$u_i=\frac{x_i-a}{h}=\frac{x_i-0.10}{0.04}$$ | fi ui |
| 0.00 – 0.04 | 4 | 0.02 | -2 | -8 |
| 0.04 – 0.08 | 9 | 0.06 | -1 | -9 |
| 0.04 – 0.08 | 9 | 0.06 | -1 | -9 |
| 0.08 – 0.12 | 9 | 0.10 = a | 0 | 0 |
| 0.12 – 0.16 | 2 | 0.14 | 1 | 2 |
| 0.16 – 0.20 | 4 | 0.18 | 2 | 8 |
| 0.20 – 0.24 | 2 | 0.22 | 3 | 6 |
| Total | N=30 | -1 |
Here, a = 0.10
h = 0.04
N = 30
$$\text{Now,\space Mean}=a+h×\frac{1}{\text{N}}×\Sigma f_iu_i\\=0.10+0.04×\frac{1}{30}×(\normalsize-1)\\=0.10=\frac{0.04}{30}$$
= 0.10 – 0.001 = 0.099 ppm
8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
| Number of days | Number of students |
| 0 – 6 | 11 |
| 6 – 10 | 10 |
| 10 – 14 | 7 |
| 14 – 20 | 4 |
| 20 – 28 | 4 |
| 28 – 38 | 3 |
| 38 – 40 | 1 |
Sol. Here, the class size varies but we will apply assumed mean method.
| Number of days | Numberof students (fi) | Class mark (xi) | di = xi – a = xi – 17 | fi di |
| 0 – 6 | 11 | 3 | -14 | -154 |
| 6 – 10 | 10 | 8 | -9 | -90 |
| 10 – 14 | 7 | 12 | -5 | -35 |
| 14 – 20 | 4 | 17=a | 0 | 0 |
| 20 – 28 | 4 | 24 | 7 | 28 |
| 28 – 38 | 3 | 33 | 16 | 48 |
| 38 – 40 | 1 | 39 | 22 | 22 |
| Total | N=40 | -181 |
Here, a = 17
N = 40
By assumed mean method,
$$\text{Mean}=a+\frac{1}{\text{N}}×\Sigma f_id_i\\=17+\frac{1}{40}×(-181)\\=17-\frac{18.1}{4}$$
= 12.48 days
9. The following tables gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
| Literacy rate (in %) | Number of cities |
| 45 – 55 | 3 |
| 55-65 | 10 |
| 65-75 | 11 |
| 75 – 85 | 8 |
| 85 – 95 | 3 |
Sol.
| Literacy rate (in %) | Number of cities (fi) | Class marks (xi) | $$u_i=\frac{x_i-a}{h}=\frac{x_i-70}{10}$$ | fi ui |
| 45 – 55 | 3 | 50 | -2 | -6 |
| 55-65 | 10 | 60 | -1 | -10 |
| 65-75 | 11 | 70=a | 0 | 0 |
| 75 – 85 | 8 | 80 | 1 | 8 |
| 85 – 95 | 3 | 90 | 2 | 6 |
| Total | N=35 | -2 |
Here, a = 70
h = 10
N = 25
By step deviation method,
$$\text{Mean}=a+h×\frac{1}{h}×\Sigma f_iu_i\\=70+10×\frac{1}{35}×(\normalsize-2)\\=70-\frac{4}{7} = 70 – 0.57 = 69.43%$$
Exercise 14.2
1. The following table shows the ages of the patients admitted in a hospital during a year.
| Age (in years) | Number of patients |
| 5 – 15 | 6 |
| 15 – 25 | 11 |
| 25 – 35 | 21 |
| 35 – 45 | 23 |
| 45 – 55 | 14 |
| 55 – 65 | 5 |
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Sol.
| Age (in years) | Number of patients (fi) | Class mark (xi) | $$u_i=\frac{x_i-a}{h}=\frac{x_i-30}{10}$$ | fi ui |
| 5 – 15 | 6 | 10 | -2 | -12 |
| 15 – 25 | 11 | 20 | -1 | -11 |
| 25 – 35 | 21 | 30=a | 0 | 0 |
| 35 – 45 | 23 | 40 | 1 | 23 |
| 45 – 55 | 14 | 50 | 2 | 28 |
| 55 – 65 | 5 | 60 | 3 | 15 |
| Total | N=80 | 43 |
Here a = 30, h = 10, N = 80
Mode :
For the given data, we have the modal class 35 – 43, with maximum frequency 23.
$$\text{Mode}=l+\begin{Bmatrix*}[r]\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix*}×h$$
Here, l = 35, fm = 23, f1 = 21, f2 = 14, h = 10
$$=35+\begin{Bmatrix*}[r]\frac{23-21}{46-21-14}\end{Bmatrix*}×10\\=35+\frac{20}{11}$$
= 35 + 1.8 = 36.8 yr
Mean :
Now, by step deviation method,
$$\text{Mean = a+h}×\frac{1}{\text{N}}×\Sigma f_iu_i\\=30+10×\frac{1}{80}×43$$
= 30 + 5.37 = 35.37 yr
Thus, mode = 36.8 yr and mean = 35.57 yr. So, we conclude that the maximum number of patients admitted in the hospital are of age 36.8 yr (approx.), whereas on an average age of a patient admitted to the hospital is 35.37 yr.
2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components
| Lifetimes (in hours) | Frequency |
| 0 – 20 | 10 |
| 20 – 40 | 35 |
| 40 – 60 | 52 |
| 60 – 80 | 61 |
| 80 – 100 | 38 |
| 100 – 120 | 29 |
Determine the modal lifetimes of the components.
Sol. Modal class of the given data is 60–80. Because it has largest frequency among the given classes of the data.
Here, (Lower limit of modal class) l = 60, fm = 61, f1 = 52, f2 = 38 and (Class width) h = 20
$$\text{Mode}=l+\begin{Bmatrix*}[r]\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix*}×h\\=60+\begin{Bmatrix*}[r]\frac{61-52}{122-52-38}\end{Bmatrix*}×20\\=60+\frac{9×20}{32}\\=60+\frac{45}{8}$$
= 60 + 5.625 = 65.625 h
3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find mean monthly expenditure.
| Expenditure (in ₹) | Number of families |
| 1000 – 1500 | 24 |
| 1500 – 2000 | 40 |
| 2000 – 2500 | 33 |
| 2500 – 3000 | 28 |
| 3000 – 3500 | 30 |
| 3500 – 4000 | 22 |
| 4000 – 4500 | 16 |
| 4500 – 5000 | 7 |
Sol. For mode : Modal class of the data is (1500-2000) because it has maximum frequency among the given classes of the data.
(Lower limit of modal class) l = 1500, fm = 40, f1 = 24, f2 = 33, (Class width) h = 500
$$\text{Mode}=l+\begin{Bmatrix*}[r]\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix*}×h\\=1500+\begin{Bmatrix*}[r]\frac{40-24}{80-24-33}\end{Bmatrix*}×500\\=1500+\frac{16×500}{23}\\=1500+\frac{8000}{23}$$
= 1500 + 347.83
= ₹ 1847.83
For mean :
| Expenditure (in ₹) | Number of families (fi) | Class mark (xi) | $$u_i=\frac{x_i-A}{h}$$ | fiui |
| 1000 – 1500 | 24 | 1250 | -3 | -72 |
| 1500 – 2000 | 40 | 1750 | -2 | -80 |
| 2000 – 2500 | 33 | 2250 | -1 | -33 |
| 2500 – 3000 | 28 | 2750=a | 0 | 0 |
| 3000 – 3500 | 30 | 3250 | 1 | 30 |
| 3500 – 4000 | 22 | 3750 | 2 | 44 |
| 4000 – 4500 | 16 | 4250 | 3 | 48 |
| 4500 – 5000 | 7 | 4750 | 4 | 28 |
| Total | 200 | -35 |
So, a = 2750, h = 500, N = 200 and Σfiui = – 35
Now, by step deviation method,
$$\text{Mean = a+h}×\frac{1}{h}×\Sigma f_iu_i\\=2750+500×\frac{1}{200}×(-35)\\=2750-\frac{5×35}{2}$$
= 2750 – 87.50 = ₹ 2662.50
Thus, the modal monthly expenditure is ₹ 1847.83 and the mean monthly expenditure is ₹ 2662.50.
4. The following distribution gives the state wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret, the two measures.
| Number of students per teacher | Number of states/UT |
| 15 – 20 | 3 |
| 20 – 25 | 8 |
| 25 – 30 | 9 |
| 30 – 35 | 10 |
| 35 – 40 | 3 |
| 40 – 45 | 0 |
| 45 – 50 | 0 |
| 50 – 55 | 2 |
Sol.
| Number of students per teacher | Number of states/UT(fi) | Class mark(xi) | $$u_i=\frac{x_i-a}{h}=\frac{x_i-32.5}{5}$$ | fi × ui |
| 15 – 20 | 3 | 17.5 | -3 | -9 |
| 20 – 25 | 8 | 22.5 | -2 | -16 |
| 25 – 30 | 9 | 27.5 | -1 | -9 |
| 30 – 35 | 10 | 32.5=a | 0 | 0 |
| 35 – 40 | 3 | 37.5 | 1 | 3 |
| 40 – 45 | 0 | 42.5 | 2 | 0 |
| 45 – 50 | 0 | 47.5 | 2 | 0 |
| 50 – 55 | 2 | 52.5 | 4 | 8 |
| Total | N=35 | -23 |
Mode : Since, largest frequency is fm = 10, so its modal class is (30–35)
∴ (Lower limit of modal class) l = 30, fm = 10, f1 = 9, f2 = 3 (Class width) h = 5
$$\text{Mode}=l+\begin{Bmatrix*}[r]\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix*}×h\\=30+\begin{Bmatrix*}[r]\frac{10-9}{20-9-3}\end{Bmatrix*}×5\\=30+\frac{5}{8}$$
= 30 + 0.625
= 30.625 = 30.6
Mean :
Here, a = 32.5, h = 5, N = 35 and Σfiui = – 23
Now, by step deviation method,
$$\text{Mean = a+h}×\frac{1}{h}×\Sigma f_iu_i\\=32.5+5×\frac{1}{35}×(-23)\\=32.5-\frac{23}{7}$$
= 32.5 – 3.3 = 29.2
Hence, mode = 30.6 and mean = 29.2. We conclude that most states/UT have a student teacher ratio of 30.6 and on an average ratio is 29.2.
5. The given distribution shows the number of runs scored by some top batsmen of the world in one day international cricket matches
| Runs scored | Number of batsmen |
| 3000 – 4000 | 4 |
| 4000 – 5000 | 18 |
| 5000 – 6000 | 9 |
| 6000 – 7000 | 7 |
| 7000 – 8000 | 6 |
| 8000 – 9000 | 3 |
| 9000 – 10000 | 1 |
| 10000 – 11000 | 1 |
Find the mode of the data.
Sol. Since, maximum frequency is fm = 18, so its modal class is 4000-5000.
∴ (Lower limit of modal class) l = 4000, fm = 18, f1 = 4, f2 = 9, (Class width) h = 1000
$$\text{Mode}=l+\begin{Bmatrix*}[r]\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix*}×h\\=4000+\begin{Bmatrix*}[r]\frac{18-4}{36-4-9}\end{Bmatrix*}×1000\\=4000+\frac{14000}{23}$$
= 4000 + 608.7
= 4608.7 runs
6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the given below.Find the mode of the data
| Number of cars | Frequency |
| 0 – 10 | 7 |
| 10 – 20 | 14 |
| 20 – 30 | 13 |
| 30 – 40 | 12 |
| 40 – 50 | 20 |
| 50 – 60 | 11 |
| 60 – 70 | 15 |
| 70 – 80 | 8 |
Sol. Since, maximum frequency is fm = 20, so its modal class is (40-50).
∴ (Lower limit of modal class) l = 40, fm = 20, f1 = 12, f2 = 11, (Class width) h = 10
$$\text{Mode = l}\space+\space\begin{Bmatrix*}[r]\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix*}×h\\=40+\begin{Bmatrix*}[r]\frac{20-12}{40-12-11}\end{Bmatrix*}×10\\=40+\frac{80}{17}$$
= 40 + 4.71
= 44.71 cars.
Exercise 14.3
1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
| Monthly consumption (in units) | Number of consumers |
| 65 – 85 | 4 |
| 85 – 105 | 5 |
| 105 – 125 | 13 |
| 125 – 145 | 20 |
| 145 – 165 | 14 |
| 165 – 185 | 8 |
| 165 – 185 | 8 |
| 185 – 205 | 4 |
Sol.
| Monthly consumption (in units) | Number of consumers (fi) | Cumulative frequency (cf) |
| 65 – 85 | 4 | 4 |
| 85 – 105 | 5 | 9 |
| 105 – 125 | 13 | cf=22 |
| 125 – 145 | 20=f | 42 |
| 145 – 165 | 14 | 56 |
| 165 – 185 | 8 | 64 |
| 185-205 | 4 | 68 |
| Total | N=68 |
$$\textbf{(i) For median} :\\\text{Here, N = 68 gives}\space\frac{N}{2}=34$$
So, we have the median class (125 – 145).
(Lower median class) l = 125, N = 68, f = 20, cf = 22, (Class width) h = 20
$$\text{Median}=l+\begin{Bmatrix*}[r]\frac{\frac{N}{2}-cf}{f}\end{Bmatrix*}×h\\=125+\begin{Bmatrix*}[r]\frac{34-22}{20}\end{Bmatrix*}×20\\=125+\frac{12}{20}×20$$
= 125 + 12 = 137 units
(ii) For mode : Since, maximum frequency is fm = 20 so its modal class is 125–145. Then f1 = 13, f2 = 14,
fm = 20, l = 125 and h = 20
$$\text{Mode}=l+\begin{Bmatrix*}[r]\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix*}×h\\=125+\begin{Bmatrix*}[r]\frac{20-13}{40-13-14}\end{Bmatrix*}×20\\=125+\frac{7×20}{13}\\=125+\frac{140}{13}$$
= 125 + 10.76 = 135.76 units
(iii) For mean
| Monthly consumption (in units) | Number of consumers (fi) | Class mark (xi) | $$u_i=\frac{x_i-a}{h}=\frac{x_i-135}{20}$$ | fi ui |
| 65 – 85 | 4 | 75 | -3 | -12 |
| 85 – 105 | 5 | 95 | -2 | -10 |
| 105 – 125 | 13 | 115 | -1 | -13 |
| 125 – 145 | 20 | a=135 | 0 | 0 |
| 145 – 165 | 14 | 155 | 1 | 14 |
| 165 – 185 | 8 | 175 | 2 | 16 |
| 185 – 205 | 4 | 195 | 3 | 12 |
| Total | N = 68 | 7 |
Here, N = 68, a = 135, h = 20 and Σfiui = 7
Now, by step deviation method
$$\text{Mean = a + h}×\frac{1}{N}×\Sigma f_iu_i\\=135+20×\frac{1}{68}×7\\=135+\frac{35}{17}$$
= 135 + 2.05 = 137.05 units
Hence, the three measures are approximately the same in this cases.
2. If the median of the distribution given below is 28.5, find the value of x and y.
| Class interval | Frequency |
| 0 – 10 | 5 |
| 10 – 20 | x |
| 20 – 30 | 20 |
| 30 – 40 | 15 |
| 40 – 50 | y |
| 50 – 60 | 5 |
| Total | 60 |
Sol.
| Class interval | Frequency (f) | Cumulative frequency |
| 0 – 10 | 5 | 5 |
| 10 – 20 | x | 5 + x = cf |
| 20 – 30 | 20 = f | 25 + x |
| 30 – 40 | 15 | 40 + x |
| 40 – 50 | y | 40 + x + y |
| 50 – 60 | 5 | 45 + x + y |
| Total | 60 |
Given, median = 28.5 lies in the class interval (20–30).
Thus, median class is (20–30).
Here, l = 20, f = 20, cf = 5 + x (Class width) h = 10, (Total observations) N = 60
$$\text{Median = l}+\begin{Bmatrix*}[r]\frac{\frac{N}{2}-cf}{f}\end{Bmatrix*}×h\\\therefore\space 28.5=20 + \begin{Bmatrix*}[r]\frac{\frac{60}{2}-(5+x)}{20}\end{Bmatrix*}×10\\\Rarr\space 8.5=\frac{25-x}{2}$$
⇒ 17 = 25 – x
⇒ x = 8
From the given table, we have
i.e., x + y + 45 = 60
⇒ x + y = 15
⇒ y = 15 – x
= 15 – 8 = 7
⇒ y = 7
Hence, the value of x and y are respectively 8 and 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 |
Sol.
| Age (in years) | Number of Policy holders | Cumulative frequency |
| Below 20 | 2=2 | 2 |
| 20 – 25 | (6 – 2) = 4 | 6 |
| 25 – 30 | (24 – 6) = 18 | 24 |
| 30 – 35 | (45 – 24) = 21 | 45 = cf |
| (Median class) 35 – 40 | (78 – 45) = 33 = f | 78 |
| 40 – 45 | (89 – 78) = 11 | 89 |
| 45 – 50 | (92 – 89) = 3 | 92 |
| 50 – 55 | (98 – 92) = 6 | 98 |
| 55 – 60 | (100 – 98) = 2 | 100 |
| Total | N = 100 |
$$\because\space\frac{N}{2}=\frac{100}{2}=50$$
Here, l = 35, f = 33, cf = 45, h = 5, N = 100
$$\text{Median = l + }\begin{Bmatrix*}[r]\frac{\frac{N}{2}-cf}{f}\end{Bmatrix*}×h\\=35+\begin{Bmatrix*}[r]\frac{50-45}{33}\end{Bmatrix*}×5\\=35+\frac{25}{33}$$
= 35 + 0.76 = 35.76 age
Hence,the median age is 35.76 years.
4. The length of 40 leaves of a plant are measured correct to the nearest millimetre and the data obtained is represented in the following table :
Find the median length of the leaves.
| Length (in mm) | Number of leaves |
| 118 – 126 | 3 |
| 127 – 135 | 5 |
| 136 – 144 | 9 |
| 145 – 153 | 12 |
| 154 – 162 | 5 |
| 163 – 171 | 4 |
| 172 – 180 | 2 |
Sol. Since, given classes is not continuous, so it needs to be converted into continuous classes for finding the median. The classes, then change to 117.5–126.5, 126.5–135.5, ... 171.5–180.5.
| Length (in mm) | Number of leaves fi | Cumulative frequency |
| 117.5 – 126.5 | 3 | 3 |
| 126.5 – 135.5 | 5 | (3 + 5) = 8 |
| 135.5 – 144.5 | 9 | (8 + 9) = 17 = cf |
| 144.5 – 153.5 | 12 = f | (17 + 12) = 29 (Median class) |
| 153.5 – 162.5 | 5 | (29 + 5) = 34 |
| 162.5 – 171.5 | 4 | (34 + 4) = 38 |
| 171.5 – 180.5 | 2 | (38 + 2) = 40 |
| Total | N = 40 |
$$\because\space\frac{N}{2}=\frac{40}{2}=20$$
Since, cumulative frequency 20 lies in the interval 144.5 – 153.5
Here, l = 144.5, f = 12, cf = 17, h = 9, N = 40
$$\text{Median = l}\space+\space\begin{Bmatrix}\frac{\frac{N}{2}-cf}{f}\end{Bmatrix} ×h\\=144.5+\begin{Bmatrix}\frac{20-17}{12}\end{Bmatrix}×9\\=144.5+\frac{3×3}{4}\\=144.5+\frac{9}{4}$$
= 144.5 + 2.25
= 146.75 mm
Hence, the median length of the leaves in 146.75 mm.
5. The following table gives the distribution of the life time of 400 neon lamps
| Life time (in hours) | Number of lamps |
| 1500 – 2000 | 14 |
| 2000 – 2500 | 56 |
| 2500 – 3000 | 60 |
| 3000 – 3500 | 86 |
| 3500 – 4000 | 74 |
| 4000 – 4500 | 62 |
| 4500 – 5000 | 48 |
Find the median life time of a lamp.
Sol.
$$\because\space\frac{N}{2}=\frac{400}{2}=200$$
Since, cumulative frequency 20 lies in the interval 3000–3500.
Here, l = 3000, f = 86, cf = 130, h = 500, N = 400
$$\text{Median = l}\space+\space\begin{Bmatrix}\frac{\frac{N}{2}-cf}{f}\end{Bmatrix}×h\\=3000+\frac{\begin{Bmatrix}200-130\end{Bmatrix}}{86}×500\\=3000+\frac{70×500}{86}\\=3000+\frac{35000}{86}$$
= 3000 + 406.98 = 3406.98 h
Hence, median life time of a lamp is 3406.98 h.
6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows :
| Number of letters | Number of surnames |
| 1 – 4 | 6 |
| 4 – 7 | 30 |
| 7 – 10 | 40 |
| 10 – 13 | 16 |
| 13 – 16 | 4 |
| 16 – 19 | 4 |
Determine the median number of letter in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Sol. (i)
| Number of letters | Number of surnames | Cumulative frequency |
| 1 – 4 | 6 | 6 |
| 4 – 7 | 30 | 6 + 30 = 36 = cf |
| 07 – 10 | 40 = f | 36 + 40 = 76 (median class) |
| 10 – 13 | 16 | 76 + 16 = 92 |
| 13 – 16 | 4 | 92 + 4 = 96 |
| 16 – 19 | 4 | 96 + 4 = 100 |
| Total | N = 100 |
$$\because\space\frac{N}{2}=\frac{100}{2}=50$$
Since, cumulative frequency 50 lies in the interval 7 – 10.
Here, l = 7, f = 40, cf = 36, h = 3, N = 100
$$\text{Median = l }+\begin{Bmatrix}\frac{\frac{N}{2}-cf}{f}\end{Bmatrix}×h\\=7+\begin{Bmatrix}\frac{50-36}{40}\end{Bmatrix}×3\\=7+\frac{14×3}{40}\\=7+\frac{21}{20}=7+1.05$$
= 8.05
Hence, the median number of letters in the surnames is 8.05.
(ii) Modal class is (7 – 10), because it has maximum frequency as fm = 40.
Here, l = 7, fm = 40, f1 = 30, f2 = 16, h = 3
$$\text{Mode = l}\space+\space\begin{Bmatrix}\frac{f_m-f_1}{2f_m-f_1-f_2}\end{Bmatrix}×h\\=7+\begin{Bmatrix}\frac{40-30}{80-30-16}\end{Bmatrix}×3\\=7+\frac{30}{34}$$
= 7 + 0.88
= 7.88
Hence, the modal size of the surnames is 7.88.
(iii)
| Number of letters | fi | Class mark (xi) | $$u_i=\frac{x_i-a}{h}=\frac{x_i-8.5}{3}$$ | fi × ui |
| 1 – 4 | 6 | 2.5 | -2 | -12 |
| 4 – 7 | 30 | 5.5 | -1 | -30 |
| 7 – 10 | 40 | 8.5 = a | 0 | 0 |
| 10 – 13 | 16 | 11.5 | 1 | 16 |
| 13 – 16 | 4 | 14.5 | 2 | 8 |
| 16 – 19 | 4 | 17.5 | 3 | 12 |
| Total | N = 100 | -6 |
Here, N = 100, a = 8.5, h = 3 and Σfiui = – 6
Now, by step deviation method,
$$\text{Mean = a+h×}\frac{1}{N}×\Sigma f_iu_i\\=8.5+3×\frac{1}{100}×(-6)\\=8.5-\frac{18}{100}$$
= 8.5 – 0.18
= 8.32
Hence, the mean number of letters in the surname is 8.32.
7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
| Weight (in kgs) | Number of students |
| 40 – 45 | 2 |
| 45 – 50 | 3 |
| 50 – 55 | 8 |
| 55 – 60 | 6 |
| 60 – 65 | 6 |
| 65 – 70 | 3 |
| 70 – 75 | 2 |
Sol.
| Weight (in kg) | Number of students (fi) | Cumulative frequency |
| 40 – 45 | 2 | 2 |
| 45 – 50 | 3 | 2+3=5 |
| 50 – 55 | 8 | 5 + 8 = 13 = cf |
| 55 – 60 | 6 = f | 13 + 6 = 19 (median class) |
| 60 – 65 | 6 | 19 + 6 = 25 |
| 65 – 70 | 3 | 25 + 3 = 28 |
| 70 – 75 | 2 | 28 + 2 = 30 |
| Total | N=30 |
$$\because\space\frac{N}{2}=\frac{30}{2}=15$$
Since, cumulative frequency 15 lies in the interval 55–60.
Here, l = 55, f = 6, cf = 13, h = 5, N = 30
$$\text{Median = l} + \begin{Bmatrix}\frac{\frac{N}{2}-cf}{f}\end{Bmatrix}×h\\55+\begin{Bmatrix}\frac{15-13}{6}\end{Bmatrix}×5\\=55+\frac{5}{3}$$
= 55 + 1.67
= 56.67 kg
Hence, the median weight of the students is 56.67 kg.
Exercise 14.4
1. The following distribution gives the daily income of 50 workers of a factory.
| Daily income (in ₹) | Number of workers |
| 100 – 120 | 12 |
| 120 – 140 | 14 |
| 140 – 160 | 8 |
| 160 – 180 | 6 |
| 180 – 200 | 10 |
Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.
Sol.
| Daily income (in ₹) | Number of workers (Frequency) | Cumulative frequency less than type | |
| 100 – 120 | 12 | Less than 120 | 12=12 |
| 120 – 140 | 14 | Less than 140 | (12 + 14) = 26 |
| 140 – 160 | 8 | Less than 160 | (26 + 8) = 34 |
| 160 – 180 | 6 | Less than 180 | (34 + 6) = 40 |
| 180 – 200 | 10 | Less than 200 | (40 + 10) = 50 |
| Total | N=50 | ||
$$\because\space\text{N = 50, gives}\space\frac{n}{2}=25$$
On the graph, we will plot the points (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50).
2. During the medical check up 35 students of a class, their weights were recorded as follows
| Weight (in kg) | Number of students |
| Less than 38 | 0 |
| Less than 40 | 3 |
| Less than 42 | 5 |
| Less than 44 | 9 |
| Less than 46 | 14 |
| Less than 48 | 28 |
| Less than 50 | 32 |
| Less than 52 | 35 |
Draw a less than ogive for the given data. Hence, obtain the median weight from the graph and verify the result by using the formula.
Sol.
| Weight (in kg) | Number of students (Frequency) fi | Cumulative frequency less than type | |
| Less than 38 | 0 = 0 | Less than 38 | 0 |
| 38 – 40 | (3 – 0) = 3 | Less than 40 | 3 |
| 40 – 42 | (5 – 3) = 2 | Less than 42 | 5 |
| 42 – 44 | (9 – 5) = 4 | Less than 44 | 9 |
| 44 – 46 | (14 – 9) = 5 | Less than 46 | 14 = cf |
| 46 – 48 | (28 – 14) = 14 = f | Less than 48 | 28 (median class) |
| 48 – 50 | (32 – 28) = 4 | Less than 50 | 32 |
| 50 – 52 | (35 – 32) = 3 | Less than 52 | 35 |
| Total | N = 35 | ||
$$\because\space\frac{N}{2}=\frac{35}{2}=17.5$$
To draw the 'less than' type ogive, we plot the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35) on the graph.
Take point 17.5 taking on y-axis draw a line parallel to x-axis meet at a point L′ and draw a perpendicular line from M′ to the x-axis, the intersection point of X-axis is the median.
Median from the graph = 46.5 kg
Median class is (46-48)
We have, l = 46, f = 14, cf = 14, h = 2, N = 35
From the table :
$$\text{Median = l} + \begin{Bmatrix}\frac{\frac{N}{2}-cf}{f}\end{Bmatrix}×h\\=46+\begin{Bmatrix}\frac{\frac{35}{2}-14}{14}\end{Bmatrix}×2\\=46+\frac{1}{2}$$
= 46 + 0.5
= 46.5 kg
Hence, the median is same as we have noticed from the graph.
3. The following table gives production yield per hectare of wheat of 100 farms of a village.
| Production yield (in kg/ha) | Number of farms |
| 50 – 55 | 2 |
| 55 – 60 | 8 |
| 60 – 65 | 12 |
| 65 – 70 | 24 |
| 70 – 75 | 38 |
| 75 – 80 | 16 |
Change the distribution to a more than type distribution and draw its ogive.
Sol.
| Production yield (in kg/ha) | Number of farms (Frequency) | Cumulative frequency more than type | |
| 50 – 55 | 2 | More than 50 | 98 + 2 = 100 |
| 55 – 60 | 8 | More than 55 | 8 + 90 = 98 |
| 60 – 65 | 12 | More than 60 | 12 + 78 = 90 |
| 65 – 70 | 24 | More than 65 | 24 + 54 = 78 |
| 70 – 75 | 38 | More than 70 | 38 + 16 = 54 |
| 75 – 80 | 16 | More than 75 | =16 |
| Total | N = 100 | ||
Now, we will draw the ogive by plotting the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54), and (75, 16).
