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Unit 14

Statistics 

 Exercise 14.1 

Q1:

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 plants0 – 22 – 44 – 66 – 88 – 1010 – 1212 – 14
Number of houses1215623

Which method did you use for finding the mean, and why?

Answer:

To find the class mark (xi) for each interval, the following relation is used.

Class mark (xi)

xi and fi(xi) can be calculated as follows.

Number of plantsNumber of houses (fi)xifi(xi)
0 – 2111 × 1 = 1
2 – 4232 × 3 = 6
4 – 6151 × 5 = 5
6 – 8575 × 7 = 35
8 – 10696 × 9 = 54
10 – 122112 ×11 = 22
12 – 143133 × 13 = 39
Total20162

From the table, it can be observed that

From the table, it can be observed that

lljfo3WYGdHLWIG1g7Hzl bFdGIV sSjRFlpsnD4L PZHxkMKWh0cjgEAFE62HXXshr jJLeVCc YqmbOMvWC kTkpwM7cMKpRCaoziwADEioYeXHG4jrHFP7qdDhf5074NXu8

Therefore, the mean daily wage of the workers of the factory is Rs 145.20.

Q3:

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.

Daily pocket allowance (in Rs)11 – 1313 – 1515 – 1717 – 1919 – 2121 – 2323 – 25
Number of workers76913f54

Answer:

To find the class mark (xi) for each interval, the following relation is used.
9ZUxfC3ZR7YCwOSi4yCDS IVEGtX FInzjp7jTJQ1caY6knLPKiHJoKQAgnZHaT8bPFS739aiaFoyicEjgOG1dm OYBq6joMEm8VyJCGjam3Td54S83a7TAjraKbVP28hiX7z9U

JhqdxTmR5hWXvoyZrVxroANEgYUKO Ft8SyNE0W BJqqtEdllaQTMCkuIYCH1n10XBTP9kEV 0Ejg4OTEiM3PVc0caSwpwGY0G81dLXB7RtWRUT4UXBlsjq8s1G2JsAW7cU DYw

Given that, mean pocket allowance,

Taking 18 as assured mean (a), di and fidi are calculated as follows.

Daily pocket allowance (in Rs)Number of childrenfiClass mark xidi = xi -18fi(di)
11 – 13712– 6– 42
13 – 15614– 4– 24
15 – 17916– 2– 18
17 – 19131800
19 – 21f2022 f
21 – 23522420
23 – 25424624
TotalXSYxaI8luAsd7iAqOATQeYgVX9PzdudzdNPNDo6Fo0gMLh4K 9IxMpVOYBhkqOQ2f – 40

From the table, we obtain
qNm ksyo0yHBGEH5J6qiSICyWKmYK5FAlNtcXnTJ0EVsA07kqBHr73zhxgYP7Kwu aFV4Vp7AEBm33XcD9G8i3Q9cwuCOu0cIvkR0rStlY 6FMSyatSob Yb5mFIUyYnyIdzm3apjHTq5xiQs2BqdweyfDEvXmPILvVBGLAqWjVJjoiqJmR e6DOvg32MP mLv71ikl3G0jSVEDLj7ZhD6tkAHBBDX5 asJu0Fa0rAwCKk mN91J A1AUwGN0k8pExKEfJjGiUF2YLIIsd3Aq4NSpMDeZ9xda0J CANntRvhyJqqkHEieaqPeqU6GJ2NEH4hHxmLhLo6tQN 8 rtKZpZZy7AgRT7AzDOqddivNBHmsHWfZVR30FubpRQ3XI

Hence, the missing frequency, f, is 20.

Q4:

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Fine the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute65 – 6868 – 7171 – 7474 – 7777 – 8080 – 8383 – 86
Number of women2438742

Answer:

To find the class mark of each interval (xi), the following relation is used.
9ZUxfC3ZR7YCwOSi4yCDS IVEGtX FInzjp7jTJQ1caY6knLPKiHJoKQAgnZHaT8bPFS739aiaFoyicEjgOG1dm OYBq6joMEm8VyJCGjam3Td54S83a7TAjraKbVP28hiX7z9U

Class size, h, of this data = 3

Taking 75.5 as assumed mean (a), di, ui, fiui are calculated as follows


Number of heart beats per minute
Number of womenfi
xi
di = xi -75.5
wXWgF0Wp1NPVkOgG8V YDhxGqWUFgiGxkJ1vX5yGG8WfuA6QCzBiUhfrI15tuOJ7BTpW2wBLYxWPCv9YARl6MvhcR77f 3S7Bly

Fiui
65 – 68266.5– 9– 3– 6
68 – 71469.5– 6– 2– 8
71 – 74372.5– 3– 1– 3
74 – 77875.5000
77 – 80778.5317
80 – 83481.5628
83 – 86284.5936
Total304

From the table, we obtain
t sIeiVkmCQML1BwhOxRe2TNsyJ58iqAjB88Vqb2Rc7zj1TUlr alqMbRq2Aa7 jNLvqGCTTjfwEzU2UryxRhFC07EX8lbkStSZg3SGXXFYUdN2hhEk8x RU4Wfz lR3rrKEB

Therefore, mean hear beats per minute for these women are 75.9 beats per minute.

*Q5:

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 mangoes50 – 5253 – 5556 – 5859 – 6162 – 64
Number of boxes1511013511525

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Answer:

Number ofBoxes fi
50 – 5215
53 – 55110
56 – 58135
59 – 61115
62 – 6425

It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore,

has to be added to the upper class limit and has to be subtracted from the lower class limit of each interval.

Class mark (xi)can be obtained by using the following relation.
lUDp8TjHnGUcaqXvkI8fuob5DPhRU4fT9v3C0NpeuylNP 8oxS74079ut1brBqcV53CZ9MFvx Im8jfwo72TbRB9QeAXhMeiyikmgm8FJgUpK95enkTECXqnIzX mmt CD R AY

Class size (h) of this data = 3

Taking 57 as assumed mean (a), di, ui, fiui are calculated as follows.

Class intervalFiXidi= xi – 57
qqumL0gf HiDNoydJZgEuIcBlPo6Lv9ZzUzz gLSBNmEiYinhNS0WdeUdH8PLt Ow8wiEpy9TqzwUvkXQEW3EdLoPY676PLC8YwmSDum VoKFNRIZly9ec4p IQfMZrVRav XA
Fiui
49.5 – 52.51551– 6– 2– 30
52.5 – 55.511054– 3– 1– 110
55.5 – 58.513557000
58.5 – 61.51156031115
61.5 – 64.525636250
Total40025

It can be observed that

Mean number of mangoes kept in a packing box is 57.19.gWWhYCF1igh7NFVUjFu4KG7M6gfQ0 KV4C8ewpEzrm3P23AM8WPtt3SiqOPD4YAHTSwPUp32FjSsSPh dC4T1RF4PiwljJnWSFwYG59dBG4i9Czf8rs jztwCIRx qFZMn4c 1M

Step deviation method is used here as the values of fi, diare big and also, there is a common

Q6:

The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs)100 – 150150 – 200200 – 250250 – 300300 – 350
Number of households451222

Find the mean daily expenditure on food by a suitable method.

Answer:

To find the class mark (xi) for each interval, the following relation is used.
DCBvrnwP9aFM7CLDQQeyFfcokAunrUQlsocos27pv6jdArgeLwUv1573XQC9Z6ACvjauEsp569KJ0VgL6LtQMB0flMs0Td7sJjIfOfJbfY7E00moQVAJ1byd0T00s7kfzReaBvc

Class size = 50

Taking 225 as assumed mean (a), di, ui, fiui are calculated as follows.

Daily expenditure (in Rs)fixiDi= xi – 2255Atv9T9ylbm2oN00ZqtxZQQXS1g4qD KcJGI8PU8ez E HSh5BsSxiWswt3 poqGa0j0xmlOVU 0ouORfiui
100 – 1504125– 100– 2– 8
150 – 2005175– 50– 1– 5
200 – 25012225000
250 – 30022755012
300 – 350232510024
Total25– 7

From the table, we obtain

1XVqP4hH2XbG28IqO43h9QipISOOQ2BmDTpG9dTgVR M2o wA8Fr5oeqd3g2kVZves6pr0pNU1Q4JoJd
gS8ioTJjpnyVF8d0tPy25q3Bf9vD bisYHBtyQJeBwE 2ZljzCeB4hE1HvRE5Ejh8k1aGjMhEFwjcNpoBZvzO9kIUB3iMMmCPq6JDfxOTdSvpqin7TgwCSuc2BKx1AYX yJaP Q

Therefore, mean daily expenditure on food is Rs 211.

*Q7:

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 – 0.044
0.04 – 0.089
0.08 – 0.129
0.12 – 0.162
0.16 – 0.204
0.20 – 0.242

Find the mean concentration of SO2 in the air.

Answer:

To find the class marks for each interval, the following relation is used.
lUDp8TjHnGUcaqXvkI8fuob5DPhRU4fT9v3C0NpeuylNP 8oxS74079ut1brBqcV53CZ9MFvx Im8jfwo72TbRB9QeAXhMeiyikmgm8FJgUpK95enkTECXqnIzX mmt CD R AY

Class size of this data = 0.04

Taking 0.14 as assumed mean (a), di, ui,fiui are calculated as follows.

Concentration of SO2   (in ppm)FrequencyFiClass markXidi= xi- 0.140Z ebC1vj0sDVWOOcrRWDwNXl2PkyyZfRuG01cafG2gD2dxiX2SU1UbOI1tNTWWz9QUqf3SNkQ CnjKdMnktZTTAF3cngPpzVTAWE2VZZy5aMisv9JBtzYmNo4Ote923kiIHggQfiui
0.00 – 0.0440.02– 0.12– 3– 12
0.04 – 0.0890.06– 0.08– 2– 18
0.08 – 0.1290.10– 0.04– 1– 9
0.12 – 0.1620.14000
0.16 – 0.2040.180.0414
0.20 – 0.2420.220.0824
Total30– 31

From the table, we obtain
vlgbdBTykVc068xODXR5rOZ6uXgR4c0z LmtlqDolpJsLgSIrLNLRB4JVyY25YQZVaImXtNvL Az5RRop5lgmUwsC6k1mlaKwVn4 We6QklHUFL KUJUbHwWh1vKouTCg9JOh3Pc8KpmQjkCrdTqGPKA3kck2INZrYICi0BIfMAYVUomUo

Therefore, mean concentration of SO2 in the air is 0.099 ppm.

Q8:

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 days0 – 66 – 1010 – 1414 – 2020 – 2828 – 3838 – 40
Number of students111074431

Answer:

To find the class mark of each interval, the following relation is used.
lUDp8TjHnGUcaqXvkI8fuob5DPhRU4fT9v3C0NpeuylNP 8oxS74079ut1brBqcV53CZ9MFvx Im8jfwo72TbRB9QeAXhMeiyikmgm8FJgUpK95enkTECXqnIzX mmt CD R AY

Taking 17 as assumed mean (a), di and fidi are calculated as follows.

Number of daysNumber of students fixidi = xi– 17fidi
0 – 6113– 14– 154
6 – 10108– 9– 90
10 – 14712– 5– 35
14 – 2041700
20 – 28424728
28 – 383331648
38 – 401392222
Total40– 181

From the table, we obtain
jXundU2KT4aIMJdyp6z94PaPBm1XJQivpeliNIVrdcRMcVDasOc0A4lCGu7or3WpVTkoQBlv3bFJt6rKq8BNTwwijEpJ iAU Bvr3x3ID8gSjvTYG3q6m1gk26 UAqNEI8bqhY

Therefore, the mean number of days is 12.48 days for which a student was absent.

Q9:

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %)45 – 5555 – 6565 – 7575 – 8585 – 95
Number of cities3101183

Answer:

To find the class marks, the following relation is used.
lUDp8TjHnGUcaqXvkI8fuob5DPhRU4fT9v3C0NpeuylNP 8oxS74079ut1brBqcV53CZ9MFvx Im8jfwo72TbRB9QeAXhMeiyikmgm8FJgUpK95enkTECXqnIzX mmt CD R AY

Class size (h) for this data = 10

Taking 70 as assumed mean (a), di, ui, and fiu  are calculated as follows.

Literacy rate (in %)Number of citiesfixidi = xi – 70fiui
45 – 55350– 20– 2– 6
55 – 651060– 10– 1– 10
65 – 751170000
75 – 858801018
85 – 953902026
Total35– 2

From the table, we obtain
h6 WhKcJHPhDBdEToVvTMRd gwaXFPndSd3WUOQP ma09wZUND12ZVevgLS7eA9G6cQs1TXR1blmV8NoJ4VA2YYHAMOlH20SGOX1BQFruT M73Zfwj2y73j2Z3He9FZDYM7X90o

Therefore, mean literacy rate is 69.43%.

 Exercise 14.2

Solutions of Questions on Page Number: 275

*Q1:

The following table shows the ages of the patients admitted in a hospital during a year:

age (in years)5 – 1515 – 2525 – 3535 – 4545 – 5555 – 65
Number of patients6112123145

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Answer:

To find the class marks (xi), the following relation is used.

Taking 30 as assumed mean (a), di and fidi are calculated as follows.lUDp8TjHnGUcaqXvkI8fuob5DPhRU4fT9v3C0NpeuylNP 8oxS74079ut1brBqcV53CZ9MFvx Im8jfwo72TbRB9QeAXhMeiyikmgm8FJgUpK95enkTECXqnIzX mmt CD R AY

Age (in years)Number of patientsfiClass markxidi= xi– 30fidi
5 – 15610– 20– 120
15 – 251120– 10– 110
25 – 35213000
35 – 45234010230
45 – 55145020280
55 – 6556030150
Total80430

From the table, we obtain
5eCMzteZ1Hke3CGc4f2WzNbQq5CVoqhN5gHZb ME3TKGwh18BdjEkNUZpJuZslsrhdqXBHZtYNW2csJer1jVaocV01qm0NN2U6P1nWgRRF5ng6A2lw 8J D9xHm5FJ7uHTgg rc

Mean of this data is 35.38. It represents that on an average, the age of a patient admitted to hospital was 35.38 years.

It can be observed that the maximum class frequency is 23 belonging to class interval 35 – 45. Modal class = 35 – 45

Lower limit (l) of modal class = 35 Frequency (f1) of modal class = 23 Class size (h) = 10

Frequency (f0) of class preceding the modal class = 21 Frequency (f2) of class succeeding the modal class = 148yVEoeAs6I6beLMUL8NX1urFUtD2EpNRffAFVjdw1mjAQWQR9 ejwG5IstrO0qditL420539JeVwoTpv Etz hoW3RGWEeik4woUverucm4MtiQdz3zfUzaBbocqq46vrTwNic

Mode =

6Dp1yZeLm88825 C9va2AIC4OpOUdq3CQ0yYvGttokWs cIOGIzGUbTR9 EuNUGlK9RPctx3 4tvwhn PcImvxNs6beF r9Dk5UAAQcPmrDfsEBpKsC6z1sn8TWImN6Il UOfaA

Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years.

Q2:

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Lifetimes (in hours)0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120
Frequency103552613829

Determine the modal lifetimes of the components.

Answer:

From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60-80.

Therefore, modal class = 60 – 80  Lower class limit (l) of modal class = 60 Frequency (f1) of modal class = 61 

Frequency (f0) of class preceding the modal class = 52 Frequency (f2) of class succeeding the 

modal class = 38 Class size (h) = 20 
LmvGBphtSGQhh6V 2tpatprvHkcIcnVXQLsvh9y oOverAE4nNgDd3yxSNUEZXqE7xTIPyijlXic6Bi AahDazjkr DAtQr eLuTcn2odpBO5b3tb9c0dYh0Ma7RZWr54TA3cac

Therefore, modal lifetime of electrical components is 65.625 hours.

Q3:

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.

Expenditure (in Rs)Number of families
1000 – 150024
1500 – 200040
2000 – 250033
2500 – 300028
3000 – 350030
3500 – 400022
4000 – 450016
4500 – 50007

Answer:

It can be observed from the given data that the maximum class frequency is 40, belonging to 1500 – 2000 intervals. Therefore, modal class = 1500 – 2000

Lower limit (l) of modal class = 1500 Frequency (f1) of modal class = 40 

Frequency (f0) of class preceding modal class = 24 Frequency (f2) of class succeeding modal class = 33 Class size (h) = 500

Therefore, modal monthly expenditure was Rs 1847.83. To find the class mark, the following relation is used.

Class size (h) of the given data = 500

Taking 2750 as assumed mean (a), di, ui, and fiui are calculated as follows.

Expenditure (in Rs)Number of familiesfixidi= xi – 2750
YWmkLpoIGjn1LBc8Vwd rRXoQeK6mY4Ww77 U4vrV7gGeojV
fiui
1000 – 1500241250– 1500– 3– 72
1500 – 2000401750– 1000– 2– 80
2000 – 2500332250– 500– 1– 33
2500 – 3000282750000
3000 – 3500303250500130
3500 – 40002237501000244
4000 – 45001642501500348
4500 – 5000747502000428
Total200– 35

From the table, we obtained

nFvQs22wDsEiKJa0EC8E7Kumqvd8esFlLyGsemo BJd vDELqQMiVdWeidBLJ6CdR9kpHrrMhY2d

EiKtZNxD33hNzbpmrrGBCWJIicQN4D7HGQAWDedtY2OVeSF jzPj9ARojdE5DM5J75hp4H28mRyC5hRhXwfQbCpM91pEvmCw3nMz0R736 hgsJ

*Q4:

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 teacherNumber of states/U.T
15 – 203
20 – 258
25 – 309
30 – 3510
35 – 403
40 – 450
45 – 500
50 – 552

Answer:

It can be observed from the given data that the maximum class frequency is 10 belonging to class interval 30 – 35. 

Therefore, 

modal class = 30 – 35

Class size (h) = 5

Lower limit (l) of modal class = 30 Frequency (f1) of modal class = 10 

Frequency (f0) of class preceding modal class = 9 Frequency (f2) of class succeeding modal class = 3

Tyre 6SqQwZJBJ FM2O7kPvxhppGnPkY ZbXNGYF10QURONBNUQPuW oYHVWQRYAr6I1AtnJnFepGBfDTqAFEfKZVJvyseCbp9yuaJN6UjJq9tEysUF5nREUDTG1uH7EQlcsO7E

It represents that most of the states/U.T have a teacher-student ratio as 30.6. To find the class marks, the following relation is used.

Yv5TF3vINeqt3 89uaLUISZMRJs14ZRJkd Ird YOwl9AVu7tquzGHat C57CbYtQuryu5yWRffX6oTp0cwSQJvj1XwRRKPJrHFQiDRlptyzM0TwwL8I8AhYv7FHA

Taking 32.5 as assumed mean (a)di, ui, fiui are calculated as follows.

Number of students per teacherNumber of states/U.T (fi)xidi = xi – 32.5
I5rYx6 O0F
fiui
15 – 20317.5– 15– 3– 9
20 – 25822.5– 10– 2– 16
25 – 30927.5– 5– 1– 9
30 – 351032.5000
35 – 40337.5513
40 – 45042.51020
45 – 50047.51530
50 – 55252.52048
Total35– 23

Therefore, mean of the data is 29.2.

Q5:

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scoredNumber of batsmen
3000 – 40004
4000 – 500018
5000 – 60009
6000 – 70007
7000 – 80006
8000 – 90003
9000 – 100001
10000 – 110001

Find the mode of the data.

Answer:

From the given data, it can be observed that the maximum class frequency is 18, belonging to class interval 4000 – 5000.

Therefore, modal class = 4000 – 5000 Lower limit (l) of modal class = 4000 Frequency (f1) of modal class = 18 

Frequency (f0) of class preceding modal class = 4 Frequency (f2) of class succeeding modal class = 9 Class size (h) = 1000

4sr9SQO2OX PlZZ wwc6mQ3W4oSgTDDhZBBUUVsTq1DgescVYeR6haGLoDrAa8sleg6 nyAzfc5UL1fb6zQxKx82x0GRcpPzAbHXW7ONPmEXtkHFFxVjKYko8DXFXhgUXbQ VAc

Therefore, mode of the given data is 4608.7 runs.

Q6:

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 table given below. Find the mode of the data:

Number of cars0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80
Frequency71413122011158

Answer:

From the given data, it can be observed that the maximum class frequency is 20, belonging to 40 – 50 class intervals.

Therefore, modal class = 40 – 50 Lower limit (l) of modal class = 40 Frequency (f1) of modal class = 20

Frequency (f0) of class preceding modal class = 12 Frequency (f2) of class succeeding modal class = 11 Class size = 10

jXSCo8BqeFl 6Ix92yTK kKl2chDz1n9M ulCWMT9FB4upvf9WYXjL7q gBRR9M43dPwUKwJ3yO38Rw1ZO

087oyGKqu1 p70WYgSrWSK9wJCkyQVHBSNG6dOu5X92Rsrkw7vj95q6Dp1xTc59Xk69y0vMU1mq1vn1oA q

Therefore, mode of this data is 44.7 cars.

Exercise 14.3 

Q1:

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 – 854
85 – 1055
105 – 12513
125 – 14520
145 – 16514
165 – 1858
185 – 2054

Answer:

To find the class marks, the following relation is used.
cWzVsyECD 3Ajt5Cd79DKFjbKiUtLIU7gtzGSf4EeYKyEYe6I hBih9mNCCmBGl63h7K481EbSEaupI iWdFmW0bEPHqNYrsKbf68t l6pFIYy ubo40gOHMCf5OC2SfAmpqsjs

Taking 135 as assumed mean (a), di, ui, fiui are calculated according to step deviation method as follows.

Monthly consumption (in units)Number of consumers (fi)xi   class markdi= xi – 135
yHhwVy0MJ7xb8

qK7Wgs6o8 XShNQzN3C9 IA6OxtGOOcqd 7KE6 04NDoZW01MLrFN6KcHDi Kx3GAHl1mLRoZVsiHi5h00Fnc61aSHdQAwFQinpd 1RcbaW3I18AzYUgl5svJQkJ3JZ yypf5B4
65 – 85475– 60– 3– 12
85 – 105595– 40– 2– 10
105 – 12513115– 20– 1– 13
125 – 14520135000
145 – 1651415520114
165 – 185817540216
185 – 205419560312
Total687

From the table, we obtain
weUPQUQZL8TMaOO6su JEgtFXVMzznOZKyiDpMiL3a8BcPz9ptZGNRMnQvE0rZ 0jYo6RWWqiWErMrN0ZIE8WrbHCYCCRLtxBFRNK WKesg1PP5gHj8Dkd9cNMVcQQYqbyNdJ8lr gKzpac5ffhhNKEntHiH1MrdS0metu6hoa9Hu9lAUw4rEWGw2qbrGoxA2H0V2 UoafsOBA

From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 – 145. Modal class = 125 – 145

Lower limit (l) of modal class = 125 Class size (h) = 20

Frequency (f1) of modal class = 20

Frequency (f0) of class preceding modal class = 13 Frequency (f2) of class succeeding the modal class

*Q2:

If the median of the distribution is given below is 28.5, find the values of x and y.

Class intervalFrequency
0 – 105
10 – 20x
20 – 3020
30 – 4015
40 – 50y
50 – 605
Total60

Answer:

The cumulative frequency for the given data is calculated as follows.

Class intervalFrequencyCumulative frequency
0 – 1055
10 – 20x5+ x
20 – 302025 + x
30 – 401540 + x
40 – 50y40+ x + y
50 – 60545 + x + y
Total (n)60

From the table, it can be observed that n = 60 45 + x + y = 60

x + y = 15 (1)

Median of the data is given as 28.5 which lies in interval 20 – 30. 

Therefore, median class = 20 – 30

Lower limit (l) of median class = 20

Cumulative frequency (cf) of class preceding the median class = 5 + x

Frequency (f) of median class = 20 Class size (h) = 10

bFEGWBxqPi7DG1eA9OT7G1L 7qTXU5SSeSP7yA1QpSBGZj5euFOU2NR523Py6heFzXy6QKfOYakR26zvGNTD35H4KAv9dVu1P1ddOuJ zTCUgFoDNds80rH1X152whCimGGDhic

From equation (1),

8 + y = 15

y = 7

Hence, the values of x and y are 8 and 7 respectively.

Q3:

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

Age (in years)Number of policy holders
Below 202
Below 256
Below 3024
Below 3545
Below 4078
Below 4589
Below 5092
Below 5598
Below 60100

Answer:

Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below.

Age (in years)Number of policy holders (fi)Cumulative frequency (cf)
18 – 2022
20 – 256 – 2 = 46
25 – 3024 – 6 = 1824
30 – 3545 – 24 = 2145
35 – 4078 – 45 = 3378
40 – 4589 – 78 = 1189
45 – 5092 – 89 = 392
50 – 5598 – 92 = 698
55 – 60100 – 98 = 2100
Total (n)

From the table, it can be observed that n = 100.

pWKesXDlUNGrWR13l9hyyKUj3rWGXuvqG FLQUoUFylJU Sl5YBUoNNywM6QUmmWG4y5kD9nxgiNI81oYBK1Wab1L8fAP Qv9kV 1NwgyxQbdWbsl 7ZZRhlcWiwt7GUFQso pQ

Cumulative frequency (cf) just greater than is 78, belonging to interval 

35 – 40. Therefore, median class = 35 – 40

Lower limit (l) of median class = 35 Class size (h) = 5

Frequency (f) of median class = 33

Cumulative frequency (cf) of class preceding median class = 45

Therefore, median age is 35.76 years.

Q4:

The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

Length (in mm)Number or leaves fi
118 – 1263
127 – 1355
136 – 1449
145 – 15312
154 – 1625
163 – 1714
172 – 1802

Find the median length of the leaves.

(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5… 171.5 – 180.5)

Answer:

The given data does not have continuous class intervals. It can be observed that the difference between two class

intervals is 1. Therefore, has to be added and subtracted to upper class limits and lower class limits respectively. 

Continuous class intervals with respective cumulative frequencies can be represented as follows.

Length (in mm)Number or leaves fiCumulative frequency
117.5 – 126.533
126.5 – 135.553 + 5 = 8
135.5 – 144.598 + 9 = 17
144.5 – 153.51217 + 12 = 29
153.5 – 162.5529 + 5 = 34
162.5 – 171.5434 + 4 = 38
171.5 – 180.5238 + 2 = 40

From the table, it can be observed that the cumulative frequency just greater 

keASljSwrBD8Q6asw0eFGLF4UY82jRfOnJr1RwOfnpS0U1bJFHRRWtCKhDPd4YV1lD66i8X2XBZRJTlMHrFXBUMftrJz8of EKvkuQUQM5wQFO4mpeO0EudlYKX3TLD8hDOwdrk

than      is 29, belonging 

to class interval 144.5 – 153.5.

Median class = 144.5 – 153.5

Lower limit (l) of median class = 144.5 Class size (h) = 9

Frequency (f) of median class = 12

Cumulative frequency (cf) of class preceding median class = 17

QAsYE9VRgNpwRAjC9Sm5F7dOLN3OwQAVcyrh0WTMIuu6CaQNZZOBDrZngAHm5WNnrHbp VqXLA1hF31QFp6MLN9ar 8eVDBls6bHUfOwm2feAZ3afLJTCAS66 IoCyfFlE4KLaI

Median

Therefore, median length of leaves is 146.75 mm.

Q5:

Find the following table gives the distribution of the life time of 400 neon lamps:

Life time (in hours)Number of lamps
1500 – 200014
2000 – 250056
2500 – 300060
3000 – 350086
3500 – 400074
4000 – 450062
4500 – 500048

Find the median life time of a lamp.

Answer:

The cumulative frequencies with their respective class intervals are as follows.

Life timeNumber of lamps (fi)Cumulative frequency
1500 – 20001414
2000 – 25005614 + 56 = 70
2500 – 30006070 + 60 = 130
3000 – 350086130 + 86 = 216
3500 – 400074216 + 74 = 290
4000 – 450062290 + 62 = 352
4500 – 500048352 + 48 = 400
Total (n)400

It can be observed that the cumulative frequency just greater than is 216, belonging to class interval 3000 – 3500.

Median class = 3000 – 3500

Lower limit (l) of median class = 3000 Frequency (f) of median class = 86

Cumulative frequency (cf) of class preceding median class = 130 Class size (h) = 500
QAsYE9VRgNpwRAjC9Sm5F7dOLN3OwQAVcyrh0WTMIuu6CaQNZZOBDrZngAHm5WNnrHbp VqXLA1hF31QFp6MLN9ar 8eVDBls6bHUfOwm2feAZ3afLJTCAS66 IoCyfFlE4KLaI

Median

8QKEcF6muScpEFlcd0ZkNVbZ6Hy1YmcQOlTjnsi8SEU06Q3 63sCUybjT18J1kpc4maG0vcR0qk2hYBazugXXDRoA69BAvC 0P5zSygOqRk Z4KUqi61m4hI MVAMwp4pJnpCYA

= 3406.976

Therefore, median life time of lamps is 3406.98 hours.

*Q6:

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 letters1 – 44 – 77 – 1010 – 1313 – 1616 – 19
Number of surnames63040644

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Answer:

The cumulative frequencies with their respective class intervals are as follows.

Number of lettersFrequency (fi)Cumulative frequency
1 – 466
4 – 73030 + 6 = 36
7 – 104036 + 40 = 76
10 – 131676 + 16 = 92
13 – 16492 + 4 = 96
16 – 19496 + 4 = 100
Total (n)100

It can be observed that the cumulative frequency just greater thanis 76, belonging to class interval 7 – 10.

Median class = 7 – 10

Lower limit (l) of median class = 7

Cumulative frequency (cf) of class preceding median class = 36 Frequency (f) of median class = 40

Class size (h) = 3
QAsYE9VRgNpwRAjC9Sm5F7dOLN3OwQAVcyrh0WTMIuu6CaQNZZOBDrZngAHm5WNnrHbp VqXLA1hF31QFp6MLN9ar 8eVDBls6bHUfOwm2feAZ3afLJTCAS66 IoCyfFlE4KLaI

Median

= 8.05

To find the class marks of the given class intervals, the following relation is used.cWzVsyECD 3Ajt5Cd79DKFjbKiUtLIU7gtzGSf4EeYKyEYe6I hBih9mNCCmBGl63h7K481EbSEaupI iWdFmW0bEPHqNYrsKbf68t l6pFIYy ubo40gOHMCf5OC2SfAmpqsjs

Taking 11.5 as assumed mean (a), di, and fiui are calculated according to step deviation method as follows.

Number of lettersNumber of surnamesfixidi = xi – 11.5
rLWEue3AX8C5OhJtiALIY1f6tXiPHnTvQptrFJezxcdsjx1QKvoDFNWu0euCWeyrioLBJNXGK ugUWCGVVKwj3MY 780hxZNyL7lxbYkCvqRlK87YRJgmv4tOCpIPz9 9t8f4 Y
fiui
1 – 462.5– 9– 3– 18
4 – 7305.5– 6– 2– 60
7 – 10408.5– 3– 1– 40
10 – 1316

Q7:

The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight (in kg)40 – 4545 – 5050 – 5555 – 6060 – 6565 – 7070 – 75
Number of students2386632

Answer:

The cumulative frequencies with their respective class intervals are as follows.

Weight (in kg)Frequency (fi)Cumulative frequency
40 – 4522
45 – 5032 + 3 = 5
50 – 5585 + 8 = 13
55 – 60613 + 6 = 19
60 – 65619 + 6 = 25
65 – 70325 + 3 = 28
70 – 75228 + 2 = 30
Total (n)30

HZyYPhBJPQfMj7f9l4XD9VZFlcRHYKKic TSWOR9ZW4BIpW91ELxvbnNkfT3NNBz1n7atdmUt5HMR2nOsl

Cumulative frequency just greater than is 19, belonging to class interval 55 – 60. 

Median class = 55 – 60

Lower limit (l) of median class = 55 Frequency (f) of median class = 6

Cumulative frequency (cf) of median class = 13 Class size (h) = 5
QAsYE9VRgNpwRAjC9Sm5F7dOLN3OwQAVcyrh0WTMIuu6CaQNZZOBDrZngAHm5WNnrHbp VqXLA1hF31QFp6MLN9ar 8eVDBls6bHUfOwm2feAZ3afLJTCAS66 IoCyfFlE4KLaI

Median
baUDqoW9y78CRGz2amT0F vG0Xh64N6eK7WxmQjtKbc2JG33n2QLPTiOiied7U6L k1iSXjzSzs3cpJtYqfQ5LbmP1IpTZ2n TjT7qP7gItt256Y05Wk0AkiFDBF4oKkGHvWuTM

= 56.67

Therefore, median weight is 56.67 kg.

Exercise 14.4 

Q1:

The following distribution gives the daily income of 50 workers of a factory.

Daily income (in Rs)100 – 120120 – 140140 – 160160 – 180180 – 200
Number of workers12148610

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Answer:

The frequency distribution table of less than type is as follows.

Daily income (in Rs) (upper class limits)Cumulative frequency
Less than 12012
Less than 14012 + 14 = 26
Less than 16026 + 8 = 34
Less than 18034 + 6 = 40
Less than 20040 + 10 = 50

Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows.
DeXwF0dJtdBJdTZ0cRCKNZcyMPndJmaJ5y1gDg0E8oO2qe57rDKhBSWtTjwoLgHhFte8DnUnyHAUl8SLl1hIP095n6GYWkOCe7hnOPGXLiprIYWtIzocL8n9j3upNza HFM1Rhc

Q2:

During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg)Number of students
Less than 380
Less than 403
Less than 425
Less than 449
Less than 4614
Less than 4828
Less than 5032
Less than 5235

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula.

Answer:

The given cumulative frequency distributions of less than types are:

Weight (in kg) upper class limitsNumber of students (cumulative frequency)
Less than 380
Less than 403
Less than 425
Less than 449
Less than 4614
Less than 4828
Less than 5032
Less than 5235

Taking upper class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be drawn as follows.

Here, n = 35

So,= 17.5

Mark the point A whose ordinate is 17.5 and its x-coordinate is 46.5. Therefore, median of this data is 46.5.

OFYf pPXkMAV2MKAYmWoR9jHGYyqzj qSzLOJNvwPAFwRyPizsDzz3KhGiRWSgYQvYxjLP98X4jsW5BRmcG9nndJb1xOQJ wSQrATl9WfLqx wcAOGlCh8EbAA84RiR0x4v9 YE

ld9eLguNHaUD 24cggyX8CDIKA NGCjulDXnbBU

It can be observed that the difference between two consecutive upper class limit is 2. The 

class marks with their respective frequencies are obtained as below.

Weight (in kg)Frequency (f)Cumulative frequency
Less than 3800
38 – 403 – 0 = 33
40 – 425 – 3 = 25
42 – 449 – 5 = 49
44 – 4614 – 9 = 514
46 – 4828 – 14 = 1428
48 – 5032 – 28 = 432
50 – 5235 – 32 = 335
Total (n)35

The cumulative frequency just greater than is 28, belonging to class interval 46 – 48. 

Median class = 46 – 48

Lower class limit (l) of median class = 46 Frequency (f) of median class = 14

Cumulative frequency (cf) of class preceding median class = 14 Class size (h) = 2

74vJd dCJm 5zWGCjtCmbfQUHk t8RslPHnCXASfSCm2y60QfkWJWmBHd1HwkN9ssZs32Nh0Ay34kNaAmuaM8bAa5XtUqlgaP0TJmhftJB2tV7g28xtISSoeXWVuy UrqxgLa 8

*Q3:

The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield (in kg/ha)50 – 5555 – 6060 – 6565 – 7070 – 7575 – 80
Number of farms2812243816

Change the distribution to a more than type distribution and draw ogive.

Answer:

The cumulative frequency distribution of more than type can be obtained as follows.

Production yield (lower class limits)Cumulative frequency
more than or equal to 50100
more than or equal to 55100 – 2 = 98
more than or equal to 6098 – 8 = 90
more than or equal to 6590 – 12 = 78
more than or equal to 7078 – 24 = 54
more than or equal to 7554 – 38 = 16

Taking the lower class limit on x-axis and their respective cumulative frequencies on y-axis, its ogive can be obtained as follows.

MdOQHoK 7zYs4vNpMYjdTqCGINbWfzQXj9JeeDGCbsgnHkJZfkt2rZ9AnrMEznjWErdcUY98apsKqP9vmnyp2XGBLPPo3E

Class 10 Maths Chapter 14, Statistics, is one of the most important of all the chapter present in the textbook. The weightage of this chapter in the final exam is around 11 to 12 marks. On average, there will be 3 questions which could be asked from this chapter and marks will be distributed in a manner of 3+4+4( it could vary as per question).

Topics covered in Chapter 14, Statistics are;

  • Mean of Grouped Data
  • Mode of Grouped Data
  • Median of Grouped Data
  • Graphical Representation of Cumulative Frequency Distribution

List of Exercises in class 10 Maths Chapter 14 :
Exercise 14.1 Solutions 9 Question ( 9 long)
Exercise 14.2 Solutions 6 Question ( 6 long)
Exercise 14.3 Solutions 7 Question ( 7 long)
Exercise 14.4 Solutions 3 Question ( 3 long)

NCERT solutions for Class 10 Maths Chapter 14- Statistics are made available for students who want to obtain good marks in this chapter. The methods and procedure to solve the questions have been explained clearly in these NCERT Solutions, such that, students find it easy to understand the fundamentals quickly.

The world is highly data-oriented, in fact, each and every field has a group of data, which represents the relevant information. Statistics is the branch of mathematics which deals with the representation of data in a meaningful way.

You will face many real-life scenarios where the fundamentals of statistics are used to represent a set of data in tabular form or in graphs or in pie charts. There are a number of methods you will learn from this chapter such as, step deviation methods, finding mode and median of grouped data, converting frequency distribution and the relation between mode, mean and median methods, etc. 10th Class NCERT solutions are the best study materials to prepare for the board exam.

Key Features of NCERT Solutions for Class 10 Maths Chapter 14- Statistics

  • The solutions for the statistics chapter works as a reference for the students.
  • It will help students to score marks against the questions asked from the statistics chapter.
  • Students can prepare and do the revision for chapter 14 with this source.
  • The questions of statistics are solved by subject experts.
  • The content of the material is as per the CBSE syllabus and guidelines.

Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 14

Does NCERT Solutions for Class 10 Maths Chapter 14 help you to clear board exams?

Yes, NCERT Solutions for Class 10 Maths Chapter 14 is one of the important chapters of Class 10 Maths of NCERT Solutions. These solutions are focused on learning various Mathematics tricks and shortcuts for quick and easy calculations. This makes them learn and clear the Maths subject in board exams.

How NCERT Solutions for Class 10 Maths Chapter 14 is helps in learning areas and volumes of geometrical shapes?

NCERT Solutions for Class 10 Maths Chapter 14 is created by our faculty in an interactive manner to make it easy for the students to understand the concepts. Students can refer to the PDF of solutions as a major study material to improve their speed in solving problems accurately.

How many questions present in NCERT Solutions for Class 10 Maths Chapter 14 ?

14th Chapter of NCERT Solutions for Class 10 Maths has 4 exercises. In that first exercise contains 9 questions, second exercise contains 6 questions, third exercise has 7 questions and last or fourth exercise has 3 questions. So there are a total of 25 long questions that are present in NCERT Solutions for Class 10 Maths Chapter 14.


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NCERT Solutions Class 10 Maths Chapters

  • Chapter 1 Real Numbers
  • Chapter 2 Polynomials
  • Chapter 3 Pair Of Linear Equations In Two Variables
  • Chapter 4 Quadratic Equations
  • Chapter 5 Arithmetic Progressions
  • Chapter 6 Triangles
  • Chapter 7 Coordinate Geometry
  • Chapter 8 Introduction To Trigonometry
  • Chapter 9 Some Applications Of Trigonometry
  • Chapter 10 Circles
  • Chapter 11 Constructions
  • Chapter 12 Areas Related To Circles
  • Chapter 13 Surface Areas And Volumes
  • Chapter 14 Statistics
  • Chapter 15 Probability

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