{"id":624192,"date":"2023-09-01T07:05:24","date_gmt":"2023-09-01T07:05:24","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=624192"},"modified":"2023-09-04T12:30:57","modified_gmt":"2023-09-04T12:30:57","slug":"kc-sinha-exercise-6-1-mathematics-solution-class-10-chapter-6-statistics","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-1-mathematics-solution-class-10-chapter-6-statistics\/","title":{"rendered":"KC Sinha: Exercise 6.1 – Mathematics Solution Class 10 Chapter 6 Statistics"},"content":{"rendered":"\n\n\n\n\n

Question 1 <\/h4>\n\n\n\n

The mean of 11 results is 30. If the mean of the first 6 results is 28 and that of last 6 results is 32, find the 6th result.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Let the 6th<\/sup> number be x
Given that mean of 11 results = 30
\u2234 sum of 11 numbers = 11 \u00d7 30 = 330
Mean of the first 6 results = 28
Sum of first 6 numbers = 6 \u00d7 28 = 168
Mean of the last 6 results = 32
Sum of the last 6 results = 6 \u00d7 32 = 192
Therefore,
Sum of first 6 numbers + sum of last 6 numbers \u2013 6th<\/sup> number = sum of 11 numbers
168 + 192 – x = 330
\u21d2 360 \u2013 x = 330
\u21d2 x = 30<\/p>\n\n\n\n

\n\n\n\n

Question 2 <\/h4>\n\n\n\n

The mean of 17 observations is 20. If the mean of the first 9 observation is 23 and that of last 9 observations is 18, find the 9th observation.<\/strong><\/p>\n\n\n\n

Sol :
Let the 9th<\/sup> observation be x
Given that mean of 17 observations = 20
\u2234 sum of 17 observations = 17 \u00d7 20 = 340
Mean of the first 9 observations= 23
Sum of first 9 observations = 9 \u00d7 23 = 207
Mean of the last 9 observations = 18
Sum of the last 9 observations = 9 \u00d7 18 = 162
Therefore,
Sum of first 9 observations + sum of last 9 observations \u2013 9th<\/sup> observation = sum of 17 observations
207 + 162 – x = 340
\u21d2 369 \u2013 x = 340
\u21d2 x = 29<\/p>\n\n\n\n

\n\n\n\n

Question 3 <\/h4>\n\n\n\n

The mean weight of 21 students of a class is 52 kg. If the mean weight of the first 11 students of the class is 50 kg and that of the last 11 students is 54 kg, find the weight of the 11th student.<\/strong><\/p>\n\n\n\n

Sol :
Let the weight of 11th<\/sup> student be x
Given that mean weight of 21 students = 52kg
\u2234 sum of 21 students weight = 21 \u00d7 52 = 1092kg
Mean weight of the first 11 students = 50kg
Sum of first 11 students weight = 11 \u00d7 50 = 550kg
Mean weight of the last 11 students = 54kg
Sum of the last 11 students weight = 11 \u00d7 54 = 594kg
Therefore,
Sum of first 11 students weight + sum of last 11 students weight \u2013 weight of the 11th<\/sup> student = sum of 21 students weight
550+594 – x = 1092
\u21d2 1144 \u2013 x = 1092
\u21d2 x = 52
Hence, weight of 11th<\/sup> student is 52kg<\/p>\n\n\n\n

\n\n\n\n

Question 4 <\/h4>\n\n\n\n

The mean weight of 25 students of a class is 60 kg. If the mean weight of the first 13 students of the class is 57 kg and that of the last 13 students is 63 kg, find the weight of the 13th student.<\/strong><\/p>\n\n\n\n

Sol :
Let the weight of 13th<\/sup> student be x
Given that mean weight of 25 students = 60kg
\u2234 sum of 25 students weight = 25 \u00d7 60 = 1500kg
Mean weight of the first 13 students = 57kg
Sum of first 13 students weight = 13 \u00d7 57 = 741kg
Mean weight of the last 13 students = 63kg
Sum of the last 13 students weight = 13 \u00d7 63 = 819kg
Therefore,
Sum of first 13 students weight + sum of last 13 students weight \u2013 weight of the 13th<\/sup> student = sum of 25 students weight
741 + 819 – x = 1500
\u21d2 1560 \u2013 x = 1500
\u21d2 x = 60
Hence, weight of 13th<\/sup> student is 60kg<\/p>\n\n\n\n

\n\n\n\n

Question 5<\/h4>\n\n\n\n

The mean of 23 observations is 34. If the mean of the first 12 observations is 32 and that of the last 12 observations is 38, find the 12th observation.<\/strong><\/p>\n\n\n\n

Sol :
Let the 12th<\/sup> observation be x
Given that mean of 23 observations = 34
\u2234 sum of 23 observations = 23 \u00d7 34 = 782
Mean of the first 12 observations = 32
Sum of first 12 observations = 12 \u00d7 32 = 384
Mean of the last 12 observations = 38
Sum of the last 12 observations = 12 \u00d7 38 = 456
Therefore,
Sum of first 12 observations + sum of last 12 observations \u2013 12th<\/sup> observation = sum of 23 observations
384 + 456 – x = 782
\u21d2 840 \u2013 x = 782
\u21d2 x = 58<\/p>\n\n\n\n

\n\n\n\n

Question 6 <\/h4>\n\n\n\n

The mean of 11 numbers is 35. If the mean of first 6 numbers is 32 and that of last 6 numbers is 37, find the 6th number.<\/strong><\/p>\n\n\n\n

Sol :
Let the 6th<\/sup> number be x
Given that mean of 11 results = 35
\u2234 sum of 11 numbers = 11 \u00d7 35 = 385
Mean of the first 6 results = 32
Sum of first 6 numbers = 6 \u00d7 32 = 192
Mean of the last 6 results = 37
Sum of the last 6 results = 6 \u00d7 37 = 222
Therefore,
Sum of first 6 numbers + sum of last 6 numbers \u2013 6th<\/sup> number = sum of 11 numbers
192 + 222 – x = 385
\u21d2 414 \u2013 x = 385
\u21d2 x = 29<\/p>\n\n\n\n

\n\n\n\n

Question 7 <\/h4>\n\n\n\n

The mean of 25 observations is 36. If the mean of the first 13 observations is 32 and that of the last 13 observations is 39, find the 13th observation.<\/strong><\/p>\n\n\n\n

Sol :
Let the 13th<\/sup> observation be x
Given that mean of 25 observations = 36
\u2234 sum of 25 observations = 25 \u00d7 36 = 900
Mean of the first 13 observations = 32
Sum of first 13 observations = 13 \u00d7 32 = 416
Mean of the last 13 observations = 39
Sum of the last 13 observations = 13 \u00d7 39 = 507
Therefore,
Sum of first 13 observations + sum of last 13 observations \u2013 13th<\/sup> observation = sum of 25 observations
416 + 507 – x = 900
\u21d2 923 \u2013 x = 900
\u21d2 x = 23<\/p>\n\n\n\n

\n\n\n\n

Question 8 <\/h4>\n\n\n\n

If the mean of the following data is 25, find the value of k.<\/strong><\/p>\n\n\n\n

x<\/td>5<\/td>15<\/td>25<\/td>35<\/td>45<\/td><\/tr>
f<\/td>3<\/td>k<\/td>3<\/td>6<\/td>2<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :<\/p>\n\n\n\n

xi<\/sub><\/td>fi<\/sub><\/td>xi<\/sub>fi<\/sub><\/td><\/tr>
5<\/td>5<\/td>15<\/td><\/tr>
15<\/td>K<\/td>15k<\/td><\/tr>
25<\/td>3<\/td>75<\/td><\/tr>
35<\/td>6<\/td>210<\/td><\/tr>
45<\/td>2<\/td>90<\/td><\/tr>
Total<\/td>\u03a3fi<\/sub>=14+k<\/td>\u03a3fi<\/sub>xi<\/sub>=390+15k<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}$
$\\Rightarrow 25=\\frac{390+15 k}{14+k}$
\u21d225(14+k) = 390+ 15k
\u21d2350 + 25k = 390 + 15k
\u21d2 25k \u2013 15k = 390 \u2013 350
\u21d2 10k = 40
\u21d2 k = 4<\/p>\n\n\n\n

\n\n\n\n

Question 9 <\/h4>\n\n\n\n

Find the arithmetic mean of the following distribution:<\/strong><\/p>\n\n\n\n

Marks obtained<\/td>10<\/td>15<\/td>20<\/td>25<\/td>30<\/td><\/tr>
No. of students<\/td>2<\/td>4<\/td>6<\/td>8<\/td>10<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :<\/p>\n\n\n\n

Marks obtained(xi<\/sub>)<\/td>No. of student(fi<\/sub>)<\/td>xi<\/sub>fi<\/sub><\/td><\/tr>
10<\/td>2<\/td>20<\/td><\/tr>
15<\/td>4<\/td>60<\/td><\/tr>
20<\/td>6<\/td>120<\/td><\/tr>
25<\/td>8<\/td>200<\/td><\/tr>
30<\/td>10<\/td>300<\/td><\/tr>
Total<\/td>\u03a3fi<\/sub>=30<\/td>\u03a3fi<\/sub>xi<\/sub>=700<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\Sigma \\mathrm{f}_{\\mathrm{i}}}=\\frac{700}{30}=23.33$<\/p>\n\n\n\n

\n\n\n\n

Question 10 <\/h4>\n\n\n\n

The mean of the following frequency distribution is 62.8. Find the missing frequency x:<\/strong><\/p>\n\n\n\n

Class<\/td>0-20<\/td>20-40<\/td>40-60<\/td>60-80<\/td>80-100<\/td>100-120<\/td><\/tr>
Frequency<\/td>5<\/td>8<\/td>x<\/td>12<\/td>7<\/td>8<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :

Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}$
$\\Rightarrow 62.8=\\frac{2640+50 \\mathrm{x}}{40+\\mathrm{x}}$
\u21d262.8 (40 + x) = 2640 + 50x
\u21d22512 + 62.8x = 2640 + 50x
\u21d2 62.8x \u2013 50x = 2640 \u2013 2512
\u21d2 12.8x = 128
\u21d2 x = 10<\/p>\n\n\n\n

\n\n\n\n

Question 11 <\/h4>\n\n\n\n

The arithmetic mean of the following data is 14. Find the value of p:<\/strong><\/p>\n\n\n\n

x<\/td>5<\/td>10<\/td>15<\/td>20<\/td>25<\/td><\/tr>
f<\/td>7<\/td>p<\/td>8<\/td>4<\/td>5<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :<\/p>\n\n\n\n

(xi<\/sub>)<\/td>(fi<\/sub>)<\/td>xi<\/sub>fi<\/sub><\/td><\/tr>
5<\/td>7<\/td>35<\/td><\/tr>
10<\/td>p<\/td>10p<\/td><\/tr>
15<\/td>8<\/td>120<\/td><\/tr>
20<\/td>4<\/td>80<\/td><\/tr>
25<\/td>5<\/td>125<\/td><\/tr>
Total<\/td>\u03a3fi<\/sub>=24+p<\/td>\u03a3fi<\/sub>xi<\/sub>=360+10p<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}$
$\\Rightarrow 14=\\frac{360+10 p}{24+p}$
\u21d214 (24+p) = 360 + 10p
\u21d2336 + 14p = 360 + 10p
\u21d2 14p \u2013 10p = 360 \u2013 336
\u21d2 4p = 24
\u21d2 p = 6<\/p>\n\n\n\n

\n\n\n\n

Question 12 <\/h4>\n\n\n\n

If the mean of the following data is 18, find the missing frequency p:<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

(xi<\/sub>)<\/td>(fi<\/sub>)<\/td>xi<\/sub>fi<\/sub><\/td><\/tr>
10<\/td>5<\/td>50<\/td><\/tr>
15<\/td>10<\/td>150<\/td><\/tr>
20<\/td>p<\/td>20p<\/td><\/tr>
25<\/td>8<\/td>200<\/td><\/tr>
Total<\/td>\u03a3fi<\/sub>=23+p<\/td>\u03a3fi<\/sub>xi<\/sub>=400+20p<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}$
$\\Rightarrow 18=\\frac{400+20 p}{23+p}$
\u21d218 (23 + p) = 400 + 20p
\u21d2414 + 18p = 400 + 20p
\u21d2 18p \u2013 20p = 400 \u2013 414
\u21d2 -2p = -14
\u21d2 p = 7<\/p>\n\n\n\n

\n\n\n\n

Question 13  <\/h4>\n\n\n\n

Find the value of p if the mean of the following distribution is 7.5:<\/strong><\/p>\n\n\n\n

x<\/td>3<\/td>5<\/td>7<\/td>9<\/td>11<\/td>13<\/td><\/tr>
f<\/td>6<\/td>8<\/td>15<\/td>p<\/td>8<\/td>4<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :

Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}$
$\\Rightarrow 7.5=\\frac{303+9 p}{41+p}$
\u21d27.5(41+p) = 303 + 9p
\u21d2307.5 + 7.5p = 303 + 9p
\u21d2 7.5p \u2013 9p = 303 \u2013 307.5
\u21d2 -1.5p = -4.5
\u21d2 p = 3<\/p>\n\n\n\n

\n\n\n\n

Question 14 <\/h4>\n\n\n\n

Find the mean of the following data:<\/strong><\/p>\n\n\n\n

Class interval<\/td>0-10<\/td>10-20<\/td>20-30<\/td>30-40<\/td>40-50<\/td><\/tr>
Frequency<\/td>12<\/td>11<\/td>8<\/td>10<\/td>9<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :

Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}=\\frac{1180}{50}=23.6$<\/p>\n\n\n\n

\n\n\n\n

Question 15 <\/h4>\n\n\n\n

Find the mean of the following distribution:<\/strong><\/p>\n\n\n\n

<\/strong>
Sol :

Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}=\\frac{1992}{100}=19.92$<\/p>\n\n\n\n

\n\n\n\n

Question 16 <\/h4>\n\n\n\n

The arithmetic mean of the following frequency distribution is 53. Find the value of p:<\/strong><\/p>\n\n\n\n

Class<\/td>0-20<\/td>20-40<\/td>40-60<\/td>60-80<\/td>80-100<\/td><\/tr>
Frequency<\/td>12<\/td>15<\/td>32<\/td>p<\/td>13<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :

Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}$
$\\Rightarrow 53=\\frac{1900+70 p}{72+p}$
\u21d253 (72+p) = 3340 + 70p
\u21d23816 + 53p = 3340 + 70p
\u21d2 53p \u2013 70p = 3340 \u2013 3816
\u21d2 -17p = -476
\u21d2 p = 28<\/p>\n\n\n\n

\n\n\n\n

Question 17 <\/h4>\n\n\n\n

If the mean of the following distribution is 5. Find the value of f1<\/sub>:<\/strong><\/p>\n\n\n\n

Class<\/td>0-20<\/td>20-40<\/td>40-60<\/td>60-80<\/td>80-100<\/td><\/tr>
Frequency<\/td>17<\/td>28<\/td>32<\/td>f1<\/sub><\/td>19<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :

Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}$
$\\Rightarrow 5=\\frac{4320+70 f_{1}}{96+f_{1}}$
\u21d2 5(96+ f1<\/sub>) = 4320 + 70f1<\/sub>
\u21d2480 + 5f1<\/sub> = 4320 + 70f1<\/sub>
\u21d2 5f1<\/sub> \u2013 70f1<\/sub> = 4320 – 480
\u21d2 -65f1<\/sub> = +3840
\u21d2 f1<\/sub> = – 59.07
This is not possible as frequency can not be negative.<\/p>\n\n\n\n

\n\n\n\n

Question 18 <\/h4>\n\n\n\n

Find the mean of the following frequency distribution:<\/strong><\/p>\n\n\n\n

Class<\/td>0-20<\/td>20-40<\/td>40-60<\/td>60-80<\/td>80-100<\/td><\/tr>
Frequency<\/td>15<\/td>18<\/td>21<\/td>29<\/td>17<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :

Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\Sigma \\mathrm{f}_{\\mathrm{i}}}=\\frac{5300}{100}=53$<\/p>\n\n\n\n

\n\n\n\n

Question 19  <\/h4>\n\n\n\n

The mean of the following frequency distribution is 62.8 and the sum of all frequency is 50. Compute the missing frequency f1<\/sub> and f2<\/sub> :<\/strong><\/p>\n\n\n\n

Class<\/td>0-20<\/td>2040<\/td>40-60<\/td>60-80<\/td>80-100<\/td>100-120<\/td>Total<\/td><\/tr>
Frequency<\/td>5<\/td>f1<\/sub><\/td>10<\/td>f2<\/sub><\/td>7<\/td>8<\/td>50<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :

Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}$
$\\Rightarrow 62.8=\\frac{2060+30 f_{1}+70 f_{2}}{50}$ [given: \u2211fi<\/sub> = 50]
\u21d2 62.8(50) = 2060 + 30f1<\/sub> +70f2<\/sub>
\u21d2 3140 = 2060 + 30f1<\/sub> +70f2<\/sub>
\u21d2 3140 – 2060 = 30f1<\/sub> +70f2<\/sub>
\u21d2 1080 = 30f1<\/sub> +70f2<\/sub>
\u21d2 108 = 3f1<\/sub> +7f2<\/sub> \u2026(i)
and 30 + f1<\/sub> +f2<\/sub> = 50
\u21d2 f1<\/sub> +f2<\/sub> = 50 \u2013 30
\u21d2 f1<\/sub> +f2<\/sub> = 20
\u21d2 f1<\/sub> = 20 \u2013 f2<\/sub> \u2026(ii)
Now, putting the value of f1<\/sub> in eq. (i), we get
3(20 \u2013f2<\/sub>) + 7f2<\/sub> = 108
\u21d2 60 \u2013 3f2<\/sub> + 7f2<\/sub> = 108
\u21d2 4f2<\/sub> = 108 \u2013 60
\u21d2 4f2<\/sub> = 48
\u21d2 f2<\/sub> = 12
Now, substitute the value of f2<\/sub> in eq. (ii), we get
f1<\/sub> = 20 \u2013 12
\u21d2 f1<\/sub> = 8<\/p>\n\n\n\n

\n\n\n\n

Question 20 <\/h4>\n\n\n\n

The mean of the following frequency distribution is 57.6 and the sum of the frequencies is 50. Find the missing frequencies f1<\/sub> and f2<\/sub> :<\/strong><\/p>\n\n\n\n

Class<\/td>0-20<\/td>20-40<\/td>40-60<\/td>60-80<\/td>80-100<\/td>100-120<\/td><\/tr>
Frequency<\/td>7<\/td>f1<\/sub><\/td>12<\/td>f2<\/sub><\/td>8<\/td>5<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :

Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{j}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}$
$\\Rightarrow 57.6=\\frac{2060+30 f_{1}+70 f_{2}}{50}$ [given: \u2211fi<\/sub> = 50]
\u21d2 57.6(50) = 1940 + 30f1<\/sub> +70f2<\/sub>
\u21d2 2880 = 1940 + 30f1<\/sub> +70f2<\/sub>
\u21d2 2880 \u2013 1940 = 30f1<\/sub> +70f2<\/sub>
\u21d2 940 = 30f1<\/sub> +70f2<\/sub>
\u21d2 94 = 3f1<\/sub> +7f2<\/sub> \u2026(i)
and 32 + f1<\/sub> +f2<\/sub> = 50
\u21d2 f1<\/sub> +f2<\/sub> = 50 \u2013 32
\u21d2 f1<\/sub> +f2<\/sub> = 18
\u21d2 f1<\/sub> = 18 \u2013 f2<\/sub> \u2026(ii)
Now, putting the value of f1<\/sub> in eq. (i), we get
3(18 \u2013f2<\/sub>) + 7f2<\/sub> = 94
\u21d2 54 \u2013 3f2<\/sub> + 7f2<\/sub> = 94
\u21d2 4f2<\/sub> = 94 \u2013 54
\u21d2 4f2<\/sub> = 40
\u21d2 f2<\/sub> = 10
Now, substitute the value of f2<\/sub> in eq. (ii), we get
f1<\/sub> = 18 \u2013 10
\u21d2 f1<\/sub> = 8<\/p>\n\n\n\n

\n\n\n\n

Question 21 <\/h4>\n\n\n\n

Find the mean of the following data:<\/strong><\/p>\n\n\n\n

Class Interval<\/td>50-60<\/td>60-70<\/td>70-80<\/td>80-90<\/td>90-100<\/td><\/tr>
Frequency<\/td>8<\/td>6<\/td>12<\/td>11<\/td>13<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :

Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}=\\frac{3900}{50}=78$<\/p>\n\n\n\n

\n\n\n\n

Question 22 <\/h4>\n\n\n\n

Find the mean of the following frequency distribution:<\/strong><\/p>\n\n\n\n

Class Interval<\/td>15-25<\/td>25-35<\/td>35-45<\/td>45-55<\/td>55-56<\/td><\/tr>
Frequency<\/td>60<\/td>35<\/td>22<\/td>18<\/td>15<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :

Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}=\\frac{4930}{150}=32.87$<\/p>\n\n\n\n

\n\n\n\n

Question 23 <\/h4>\n\n\n\n

Find the mean of the following frequency distribution:<\/strong><\/p>\n\n\n\n

Class<\/td>50-60<\/td>70-90<\/td>90-110<\/td>110-130<\/td>130-150<\/td>150-170<\/td><\/tr>
Frequency<\/td>18<\/td>12<\/td>13<\/td>27<\/td>8<\/td>22<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :
Here, the class size varies, and xi<\/sub>\u2019s are large. Now, we apply the step deviation method with a = 120 and h = 20

Now, $\\overline{\\mathrm{x}}=\\mathrm{a}+\\mathrm{h}\\left(\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{u}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}\\right)$
$\\Rightarrow \\overline{\\mathrm{x}}=120+20\\left(\\frac{-39}{100}\\right)$
$\\Rightarrow \\overline{\\mathrm{x}}=120+\\left(\\frac{-78}{10}\\right)$
$\\Rightarrow \\overline{\\mathrm{x}}=\\frac{1200-78}{10}$
\u21d2 $\\overline{\\mathrm{x}}$= 112.2<\/p>\n\n\n\n

\n\n\n\n

Question 24 <\/h4>\n\n\n\n

Find the mean of the following frequency distribution:<\/strong><\/p>\n\n\n\n

Class<\/td>25-29<\/td>30-34<\/td>35-39<\/td>40-44<\/td>45-49<\/td>50-54<\/td>55-59<\/td><\/tr>
Frequency<\/td>14<\/td>22<\/td>16<\/td>6<\/td>5<\/td>3<\/td>4<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :
Here, we can see that the class interval is not continuous. So, we make it continuous.

Now, $\\overline{\\mathrm{x}}=\\mathrm{a}+\\mathrm{h}\\left(\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{u}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}\\right)$
$\\Rightarrow \\overline{\\mathrm{x}}=42+5\\left(\\frac{-79}{70}\\right)$
$\\Rightarrow \\bar{x}=\\frac{2940-395}{70}$
\u21d2 x\u0304 = 36.36<\/p>\n\n\n\n

\n\n\n\n

Question 25 <\/h4>\n\n\n\n

The following table gives the marks scored by 50 students in a class-test:<\/strong><\/p>\n\n\n\n

Marks<\/td>0-100<\/td>100-200<\/td>200-300<\/td>300-400<\/td>400-500<\/td>500-600<\/td><\/tr>
No. of students<\/td>2<\/td>8<\/td>12<\/td>20<\/td>5<\/td>3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Find the mean marks scored by a student in the class-test.<\/strong>
Sol :
Here, the xi<\/sub>\u2019s are large. Now, we apply the step deviation method with a = 350 and h = 100

Now, $\\overline{\\mathrm{x}}=\\mathrm{a}+\\mathrm{h}\\left(\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{u}_{\\mathrm{i}}}{\\Sigma \\mathrm{f}_{\\mathrm{i}}}\\right)$
$\\Rightarrow \\bar{x}=350+100\\left(\\frac{-23}{50}\\right)$
\u21d2x\u0304 = 350 – 46
\u21d2x\u0304 = 304
Hence, the mean marks scored by a student in the class-test is 304<\/p>\n\n\n\n

\n\n\n\n

Question 26 <\/h4>\n\n\n\n

Find the mean of the following data:<\/strong><\/p>\n\n\n\n

Class Interval<\/td>0-10<\/td>10-20<\/td>20-30<\/td>30-40<\/td>40-50<\/td><\/tr>
Frequency<\/td>3<\/td>5<\/td>9<\/td>5<\/td>3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :<\/p>\n\n\n\n


Now, $\\overline{\\mathrm{x}}=\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}=\\frac{625}{25}=25$<\/p>\n\n\n\n

\n\n\n\n

Question 27 <\/h4>\n\n\n\n

Find the mean of the following data:<\/strong><\/p>\n\n\n\n

Class Interval<\/td>0-100<\/td>100-200<\/td>200-300<\/td>300-400<\/td>400-500<\/td><\/tr>
Frequency<\/td>6<\/td>9<\/td>15<\/td>12<\/td>8<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Sol :
Here, the xi<\/sub>\u2019s are large. Now, we apply the step deviation method with a = 250 and h = 100

Now, $\\overline{\\mathrm{x}}=\\mathrm{a}+\\mathrm{h}\\left(\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{u}_{\\mathrm{i}}}{\\sum \\mathrm{f}_{\\mathrm{i}}}\\right)$
$\\Rightarrow \\overline{\\mathrm{x}}=250+100\\left(\\frac{7}{50}\\right)$
\u21d2x\u0304 = 250 + 14
\u21d2x\u0304 = 264<\/p>\n\n\n\n

\n\n\n\n

Question 28 <\/h4>\n\n\n\n

The following table gives the marks scored by 80 students in a class-test:<\/strong><\/p>\n\n\n\n

Marks<\/td>0-50<\/td>50-100<\/td>100-150<\/td>150-200<\/td>200-250<\/td>250-300<\/td><\/tr>
No. of students<\/td>8<\/td>12<\/td>20<\/td>25<\/td>10<\/td>5<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Find the mean marks scored by a student in the class-test.<\/strong>
Sol :
Here, the xi<\/sub>\u2019s are large. Now, we apply the step deviation method with a = 175 and h = 50

Now, $\\overline{\\mathrm{x}}=\\mathrm{a}+\\mathrm{h}\\left(\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{u}_{\\mathrm{i}}}{\\Sigma \\mathrm{f}_{\\mathrm{i}}}\\right)$
$\\Rightarrow \\bar{x}=175+50\\left(\\frac{-48}{80}\\right)$
\u21d2x\u0304 = 175 – 30
\u21d2x\u0304 = 145
Hence, the mean marks scored by a student in the class-test is 145<\/p>\n\n\n\n

\n\n\n\n

Question 29 <\/h4>\n\n\n\n

The following table gives the distribution of expenditure of different families on education. Find the mean expenditure on education of a family:<\/strong><\/p>\n\n\n\n

Expenditure
(in Rs)<\/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

Sol :
Here, the xi<\/sub>\u2019s are large. Now, we apply the step deviation method with a = 3250 and h = 500

Now, $\\overline{\\mathrm{x}}=\\mathrm{a}+\\mathrm{h}\\left(\\frac{\\sum \\mathrm{f}_{\\mathrm{i}} \\mathrm{u}_{\\mathrm{i}}}{\\Sigma \\mathrm{f}_{\\mathrm{i}}}\\right)$
$\\Rightarrow \\bar{x}=3250+500\\left(\\frac{-235}{200}\\right)$
\u21d2x\u0304 = 3250 \u2013 587.5
\u21d2x\u0304 = 2662.5
Hence, the mean expenditure on education of a family is Rs 2662.5<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 10 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Question 1  The mean of 11 results is 30. If the mean of the first 6 results is 28 and that of last 6 results is 32, find the 6th result. Sol : Let the 6th number be xGiven that mean of 11 results = 30\u2234 sum of 11 numbers = 11 \u00d7 30 = 330Mean of […]<\/p>\n","protected":false},"author":302,"featured_media":624228,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[24],"tags":[],"boards":[],"class_list":["post-624192","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-10","entry"],"yoast_head":"\nKC Sinha: Exercise 6.1 - Mathematics Solution Class 10 Chapter 6 Statistics - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 1  The mean of 11 results is 30. If the mean of the first 6 results is 28 and that of last 6 results is 32, find the 6th result. 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Sol : Let\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-1-mathematics-solution-class-10-chapter-6-statistics\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" content=\"2023-09-01T07:05:24+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-09-04T12:30:57+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-17-scaled.jpg\" \/>\n\t<meta property=\"og:image:width\" content=\"1600\" \/>\n\t<meta property=\"og:image:height\" content=\"901\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"Pooja\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Pooja\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"15 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-1-mathematics-solution-class-10-chapter-6-statistics\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-1-mathematics-solution-class-10-chapter-6-statistics\/\"},\"author\":{\"name\":\"Pooja\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\"},\"headline\":\"KC Sinha: Exercise 6.1 – Mathematics Solution Class 10 Chapter 6 Statistics\",\"datePublished\":\"2023-09-01T07:05:24+00:00\",\"dateModified\":\"2023-09-04T12:30:57+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-1-mathematics-solution-class-10-chapter-6-statistics\/\"},\"wordCount\":2248,\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-1-mathematics-solution-class-10-chapter-6-statistics\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-17-scaled.jpg\",\"articleSection\":[\"Class 10\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-1-mathematics-solution-class-10-chapter-6-statistics\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-1-mathematics-solution-class-10-chapter-6-statistics\/\",\"name\":\"KC Sinha: Exercise 6.1 - Mathematics Solution Class 10 Chapter 6 Statistics - IndCareer Schools\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-1-mathematics-solution-class-10-chapter-6-statistics\/#primaryimage\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-6-1-mathematics-solution-class-10-chapter-6-statistics\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-17-scaled.jpg\",\"datePublished\":\"2023-09-01T07:05:24+00:00\",\"dateModified\":\"2023-09-04T12:30:57+00:00\",\"description\":\"Question 1  The mean of 11 results is 30. 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