{"id":55513,"date":"2020-11-26T14:09:03","date_gmt":"2020-11-26T14:09:03","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=55513"},"modified":"2023-09-19T02:48:05","modified_gmt":"2023-09-19T02:48:05","slug":"ncert-solutions-for-chapter-14-statistics","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-14-statistics\/","title":{"rendered":"NCERT Solutions for Maths: Chapter 14 – Statistics"},"content":{"rendered":"\n

Class 10: Maths Chapter 14 solutions. Complete Class 10 Maths Chapter 14 Notes.<\/p>\n\n\n\n

NCERT Solutions for Maths: Chapter 14 – Statistics<\/strong><\/h2>\n\n\n\n

NCERT 10th Maths Chapter 14, class 10 Maths Chapter 14 solutions<\/p>\n\n\n\n

Page No: 270<\/p>\n\n\n\n

Exercise 14.1<\/h5>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

Number of Plants<\/td>0-2<\/td>2-4<\/td>4-6<\/td>6-8<\/td>8-10<\/td>10-12<\/td>12-14<\/td><\/tr>
Number of Houses <\/td>1<\/td>2<\/td>1<\/td>5<\/td>6<\/td>2<\/td>3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Which method did you use for finding the mean, and why?<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

No. of plants
   (Class interval)  <\/td>		No. of houses (fi<\/sub>)<\/td>Mid-point (xi<\/sub>)<\/td>    fi<\/sub>xi    <\/sub><\/td><\/tr>
0-2<\/td>1<\/td>1<\/td>1<\/td><\/tr>
2-4<\/td>2<\/td>3<\/td>6<\/td><\/tr>
4-6<\/td>1<\/td>5<\/td>5<\/td><\/tr>
6-8<\/td>5<\/td>7<\/td>35<\/td><\/tr>
8-10<\/td>6<\/td>9<\/td>54<\/td><\/tr>
10-12<\/td>2<\/td>11<\/td>22<\/td><\/tr>
12-14<\/td>3<\/td>13<\/td>39<\/td><\/tr>
<\/td>Sum fi <\/sub>= 20<\/td><\/td>   Sum fi<\/sub>xi<\/sub> = 162    <\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Mean = x\u0304 = \u2211fi<\/sub>xi<\/sub> \/\u2211fi <\/sub>= 162\/20 = 8.1
We would use direct method because the numerical value of fi<\/sub> and xi<\/sub> are small.<\/p>\n\n\n\n

2. Consider the following distribution of daily wages of 50 workers of a factory.<\/strong><\/p>\n\n\n\n

Daily wages (in Rs.)<\/td>   100-120   <\/td>    120-140   <\/td>   140-160   <\/td>   160-180   <\/td>    180-200   <\/td><\/tr>
Number of workers    <\/td>12<\/td>14<\/td>8<\/td>6<\/td>10<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Find the mean daily wages of the workers of the factory by using an appropriate method.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Here, the value of mid-point (xi<\/sub>) is very large, so assumed mean A = 150 and class interval is h = 20.
So, ui <\/sub>= (xi<\/sub> – A)\/h = ui <\/sub>= (xi<\/sub> – 150)\/20<\/p>\n\n\n\n

Daily wages
   (Class interval)  <\/td>		Number of workers
frequency (fi<\/sub>)<\/td>		Mid-point (xi<\/sub>)<\/td>ui <\/sub>= (xi<\/sub> – 150)\/20<\/td>    fi<\/sub>ui    <\/sub><\/td><\/tr>
100-120<\/td>12<\/td>110<\/td>-2<\/td>-24<\/td><\/tr>
120-140<\/td>14<\/td>130<\/td>-1<\/td>-14<\/td><\/tr>
140-160<\/td>8<\/td>150<\/td>0<\/td>0<\/td><\/tr>
160-180<\/td>6<\/td>170<\/td>1<\/td>6<\/td><\/tr>
180-200<\/td>10<\/td>190<\/td>2<\/td>20<\/td><\/tr>
Total<\/td>Sum fi <\/sub>= 50<\/td><\/td><\/td>Sum fi<\/sub>ui<\/sub> = -12  <\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Mean = x\u0304 = A + h\u2211fi<\/sub>ui<\/sub> \/\u2211fi <\/sub>=150 + (20 \u00d7 -12\/50) = 150 – 4.8 = 145.20
Thus, mean daily wage = Rs. 145.20<\/p>\n\n\n\n

NCERT 10th Maths Chapter 14, class 10 Maths Chapter 14 solutions<\/p>\n\n\n\n

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Here, the value of mid-point (xi<\/sub>)  mean x\u0304 = 18<\/p>\n\n\n\n

Class interval<\/td>Number of children (fi<\/sub>)<\/td>Mid-point (xi<\/sub>)<\/td>    fi<\/sub>xi    <\/sub><\/td><\/tr>
11-13<\/td>7<\/td>12<\/td>84<\/td><\/tr>
13-15<\/td>6<\/td>14<\/td>84<\/td><\/tr>
15-17<\/td>9<\/td>16<\/td>144<\/td><\/tr>
17-19<\/td>13<\/td>18 = A<\/td>234<\/td><\/tr>
19-21<\/td>f<\/td>20<\/td>20f<\/td><\/tr>
21-23<\/td>5<\/td>22<\/td>110<\/td><\/tr>
23-25<\/td>4<\/td>24<\/td>96<\/td><\/tr>
Total<\/td>fi<\/sub> = 44+f<\/td><\/td>Sum fi<\/sub>xi<\/sub> = 752+20f <\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Mean = x\u0304 = \u2211fi<\/sub>xi<\/sub> \/\u2211fi <\/sub>= (752+20f)\/(44+f)
\u21d2 18 = (752+20f)\/(44+f)
\u21d2 18(44+f) = (752+20f)
\u21d2 792+18f = 752+20f
\u21d2 792+18f = 752+20f
\u21d2 792 – 752 = 20f – 18f
\u21d2 40 = 2f
\u21d2 f = 20<\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

xi <\/sub>= (Upper limit + Lower limit)\/2
Class size (h) = 3
Assumed mean (A) = 75.5<\/p>\n\n\n\n

Class Interval<\/td>Number of women (fi<\/sub>)<\/td>Mid-point (xi<\/sub>)<\/td>ui<\/sub> = (xi<\/sub> – 75.5)\/h<\/td>fi<\/sub>ui<\/sub><\/td><\/tr>
65-68<\/td>2<\/td>66.5<\/td>-3<\/td>-6<\/td><\/tr>
68-71<\/td>4<\/td>69.5<\/td>-2<\/td>-8<\/td><\/tr>
71-74<\/td>3<\/td>72.5<\/td>-1<\/td>-3<\/td><\/tr>
74-77<\/td>8<\/td>75.5<\/td>0<\/td>0<\/td><\/tr>
77-80<\/td>7<\/td>78.5<\/td>1<\/td>7<\/td><\/tr>
80-83<\/td>4<\/td>81.5<\/td>3<\/td>8<\/td><\/tr>
83-86<\/td>2<\/td>84.5<\/td>3<\/td>6<\/td><\/tr>
<\/td>Sum fi<\/sub>= 30<\/td><\/td><\/td>Sum fi<\/sub>ui <\/sub>= 4<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Mean = x\u0304 = A + h\u2211fi<\/sub>ui<\/sub> \/\u2211fi <\/sub>= 75.5 + 3\u00d7(4\/30) = 75.5 + 4\/10 = 75.5 + 0.4 = 75.9
The mean heart beats per minute for these women is 75.9<\/p>\n\n\n\n

NCERT 10th Maths Chapter 14, class 10 Maths Chapter 14 solutions<\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit.
Here, assumed mean (A) = 57
Class size (h) = 3<\/p>\n\n\n\n

Class Interval<\/td>Number of boxes (fi<\/sub>)<\/td>Mid-point (xi<\/sub>)<\/td>di<\/sub> = xi<\/sub> – A<\/td>fi<\/sub>di<\/sub><\/td><\/tr>
49.5-52.5<\/td>15<\/td>51<\/td>-6<\/td>90<\/td><\/tr>
52.5-55.5<\/td>110<\/td>54<\/td>-3<\/td>-330<\/td><\/tr>
55.5-58.5<\/td>135<\/td>57 = A<\/td>0<\/td>0<\/td><\/tr>
58.5-61.5<\/td>115<\/td>60<\/td>3<\/td>345<\/td><\/tr>
61.5-64.5<\/td>25<\/td>63<\/td>6<\/td>150<\/td><\/tr>
<\/td>Sum fi<\/sub> = 400<\/td><\/td><\/td>Sum fi<\/sub>di<\/sub> = 75<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Mean = x\u0304 = A + \u2211fi<\/sub>di<\/sub> \/\u2211fi <\/sub>= 57 + (75\/400) = 57 + 0.1875 = 57.19<\/p>\n\n\n\n

6. The table below shows the daily expenditure on food of 25 households in a locality.<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Find the mean daily expenditure on food by a suitable method.<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Here, assumed mean (A) = 225<\/p>\n\n\n\n

Class Interval<\/td>Number of households (fi<\/sub>)<\/td>Mid-point (xi<\/sub>)<\/td>di<\/sub> = xi<\/sub> – A<\/td>fi<\/sub>di<\/sub><\/td><\/tr>
100-150<\/td>4<\/td>125<\/td>-100<\/td>-400<\/td><\/tr>
150-200<\/td>5<\/td>175<\/td>-50<\/td>-250<\/td><\/tr>
200-250<\/td>12<\/td>225<\/td>0<\/td>0<\/td><\/tr>
250-300<\/td>2<\/td>275<\/td>50<\/td>100<\/td><\/tr>
300-350<\/td>2<\/td>325<\/td>100<\/td>200<\/td><\/tr>
<\/td>Sum fi<\/sub> = 25<\/td><\/td><\/td>Sum fi<\/sub>di<\/sub> = -350<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Mean = x\u0304 = A + \u2211fi<\/sub>di<\/sub> \/\u2211fi <\/sub>= 225 + (-350\/25) = 225 – 14 = 211
The mean daily expenditure on food is 211<\/p>\n\n\n\n

NCERT 10th Maths Chapter 14, class 10 Maths Chapter 14 solutions<\/p>\n\n\n\n

7. To find out the concentration of SO2<\/sub> 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:<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Find the mean concentration of SO2<\/sub> in the air.<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Concentration of SO2 <\/sub>(in ppm)<\/td>Frequency (fi<\/sub>)<\/td>Mid-point (xi<\/sub>)<\/td>fi<\/sub>xi<\/sub><\/td><\/tr>
0.00-0.04<\/td>4<\/td>0.02<\/td>0.08<\/td><\/tr>
0.04-0.08<\/td>9<\/td>0.06<\/td>0.54<\/td><\/tr>
0.08-0.12<\/td>9<\/td>0.10<\/td>0.90<\/td><\/tr>
0.12-0.16<\/td>2<\/td>0.14<\/td>0.28<\/td><\/tr>
0.16-0.20<\/td>4<\/td>0.18<\/td>0.72<\/td><\/tr>
0.20-0.24<\/td>2<\/td>0.20<\/td>0.40<\/td><\/tr>
Total<\/td>Sum fi<\/sub> = 30<\/td><\/td>Sum (fi<\/sub>xi<\/sub>) = 2.96<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Mean = x\u0304 = \u2211fi<\/sub>xi<\/sub> \/\u2211fi<\/sub>
= 2.96\/30 = 0.099 ppm<\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

Number of days<\/th>0-6<\/td>6-10<\/td>10-14<\/td>14-20<\/td>20-28<\/td>28-38<\/td>38-40<\/td><\/tr>
Number of students<\/th>11<\/td>10<\/td>7<\/td>4<\/td>4<\/td>3<\/td>1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Class interval<\/td>Frequency (fi<\/sub>)<\/td>Mid-point (xi<\/sub>)<\/td>fi<\/sub>xi<\/sub><\/td><\/tr>
0-6<\/td>11<\/td>3<\/td>33<\/td><\/tr>
6-10<\/td>10<\/td>8<\/td>80<\/td><\/tr>
10-14<\/td>7<\/td>12<\/td>84<\/td><\/tr>
14-20<\/td>4<\/td>17<\/td>68<\/td><\/tr>
20-28<\/td>4<\/td>24<\/td>96<\/td><\/tr>
28-38<\/td>3<\/td>33<\/td>99<\/td><\/tr>
38-40<\/td>1<\/td>39<\/td>39<\/td><\/tr>
<\/td>Sum fi<\/sub> = 40<\/td><\/td>Sum fi<\/sub>xi<\/sub> = 499<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Mean = x\u0304 = \u2211fi<\/sub>xi<\/sub> \/\u2211fi<\/sub>
= 499\/40 = 12.48 days<\/p>\n\n\n\n

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
<\/strong><\/p>\n\n\n\n

Literacy rate (in %)<\/td>45-55<\/td>55-65<\/td>65-75<\/td>75-85<\/td>85-98<\/td><\/tr>
Number of cities<\/td>3<\/td>10<\/td>11<\/td>8<\/td>3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Class Interval<\/td>Frequency (fi<\/sub>)<\/td>(xi<\/sub>)<\/td>di<\/sub> = xi<\/sub> – a<\/td>ui<\/sub> = di<\/sub>\/h<\/td>fi<\/sub>ui<\/sub><\/td><\/tr>
45-55<\/td>3<\/td>50<\/td>-20<\/td>-2<\/td>-6<\/td><\/tr>
55-65<\/td>10<\/td>60<\/td>-10<\/td>-1<\/td>-10<\/td><\/tr>
65-75<\/td>11<\/td>70<\/td>0<\/td>0<\/td>0<\/td><\/tr>
75-85<\/td>8<\/td>80<\/td>10<\/td>1<\/td>8<\/td><\/tr>
85-95<\/td>3<\/td>90<\/td>20<\/td>2<\/td>6<\/td><\/tr>
<\/td>Sum fi<\/sub>  = 35<\/td><\/td><\/td><\/td>Sum fi<\/sub>ui<\/sub>  = -2<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Mean = x\u0304 = a + (\u2211fi<\/sub>ui<\/sub> \/\u2211fi<\/sub>) \u0445 h
= 70 + (-2\/35) \u0445 10 = 69.42<\/p>\n\n\n\n

Exercise 14.2<\/h2>\n\n\n\n

1. The following table shows the ages of the patients admitted in a hospital during a year:
<\/strong><\/p>\n\n\n\n

Age (in years)<\/td>5-15<\/td>15-25<\/td>25-35<\/td>35-45<\/td>45-55<\/td>55-65<\/td><\/tr>
Number of patients<\/td>6<\/td>11<\/td>21<\/td>23<\/td>14<\/td>5<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n


Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Modal class = 35 \u2013 45, l = 35, class width (h) = 10, fm<\/sub> = 23, f1<\/sub> = 21 and f2<\/sub> = 14<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Calculation of Mean:<\/p>\n\n\n\n

Class Interval<\/td>Frequency (fi<\/sub>)<\/td>Mid-point (xi<\/sub>)<\/td>fi<\/sub>xi<\/sub><\/td><\/tr>
5-15<\/td>6<\/td>10<\/td>60<\/td><\/tr>
15-25<\/td>11<\/td>20<\/td>220<\/td><\/tr>
25-35<\/td>21<\/td>30<\/td>630<\/td><\/tr>
35-45<\/td>23<\/td>40<\/td>920<\/td><\/tr>
45-55<\/td>14<\/td>50<\/td>700<\/td><\/tr>
55-65<\/td>5<\/td>60<\/td>300<\/td><\/tr>
<\/td>Sum fi<\/sub> = 80<\/td><\/td>Sum fi<\/sub>xi<\/sub> = 2830<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Mean = x\u0304 = \u2211fi<\/sub>xi<\/sub> \/\u2211fi<\/sub>
= 2830\/80 = 35.37 yr<\/p>\n\n\n\n

NCERT 10th Maths Chapter 14, class 10 Maths Chapter 14 solutions<\/p>\n\n\n\n

2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :
<\/strong><\/p>\n\n\n\n

Lifetime (in hours)<\/td>0-20<\/td>20-40<\/td>40-60<\/td>60-80<\/td>80-100<\/td>100-120<\/td><\/tr>
Frequency<\/td>10<\/td>35<\/td>52<\/td>61<\/td>38<\/td>29<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n


Determine the modal lifetimes of the components.<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Modal class of the given data is 60\u201380.
Modal class = 60-80, l = 60, fm<\/sub> = 61, f1<\/sub> = 52, f2<\/sub> = 38 and h = 20<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

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 the mean monthly expenditure :
<\/strong><\/p>\n\n\n\n

Expenditure<\/td>Number of families<\/td><\/tr>
1000-1500<\/td>24<\/td><\/tr>
1500-2000<\/td>40<\/td><\/tr>
2000-2500<\/td>33<\/td><\/tr>
2500-3000<\/td>28<\/td><\/tr>
3000-3500<\/td>30<\/td><\/tr>
3500-4000<\/td>22<\/td><\/tr>
4000-4500<\/td>16<\/td><\/tr>
4500-5000<\/td>7<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Modal class = 1500-2000, l = 1500, fm<\/sub> = 40, f1<\/sub> = 24, f2<\/sub> = 33 and h = 500<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Calculation for mean:<\/p>\n\n\n\n

Class Interval<\/td>fi<\/td>xi<\/td>di = xi – a<\/td>ui = di\/h<\/td>fiui<\/td><\/tr>
1000-1500<\/td>24<\/td>1250<\/td>-1500<\/td>-3<\/td>-72<\/td><\/tr>
1500-2000<\/td>40<\/td>1750<\/td>-1000<\/td>-2<\/td>-80<\/td><\/tr>
2000-2500<\/td>33<\/td>2250<\/td>-500<\/td>-1<\/td>-33<\/td><\/tr>
2500-3000<\/td>28<\/td>2750<\/td>0<\/td>0<\/td>0<\/td><\/tr>
3000-3500<\/td>30<\/td>3250<\/td>500<\/td>1<\/td>30<\/td><\/tr>
3500-4000<\/td>22<\/td>3750<\/td>1000<\/td>2<\/td>44<\/td><\/tr>
4000-4500<\/td>16<\/td>4250<\/td>1500<\/td>3<\/td>48<\/td><\/tr>
4500-5000<\/td>7<\/td>4750<\/td>2000<\/td>4<\/td>28<\/td><\/tr>
<\/td>fi = 200<\/td><\/td><\/td><\/td>fiui = -35<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Mean = x\u0304 = a + (\u2211fi<\/sub>ui<\/sub> \/\u2211fi<\/sub>) \u0445 h
= 2750 + (35\/200) \u0445 500
= 2750 – 87.50 = 2662.50<\/p>\n\n\n\n

NCERT 10th Maths Chapter 14, class 10 Maths Chapter 14 solutions<\/p>\n\n\n\n

NCERT Solutions for Maths: Chapter 14: Download PDF<\/strong><\/h2>\n\n\n\n

NCERT Solutions for Maths: Chapter 14 – Statistics<\/p>\n\n\n\n

Download PDF<\/strong>: NCERT Solutions for Maths: Chapter 14 – Statistics PDF<\/a><\/p>\n\n\n\n

Chapterwise NCERT Solutions for Class 10 Maths<\/strong>:<\/h2>\n\n\n\n