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NCERT Solutions Class 10 Maths Chapter 14 Statistics – Here are all the NCERT solutions for Class 10 Maths Chapter 14. This solution contains questions, answers, images, explanations of the complete Chapter 14 titled Statistics of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 14 Statistics. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 14 Statistics in one place.

NCERT Solutions Class 9 Maths Chapter 14 Statistics

Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 9 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 14 Statistics , Maths, Class 9.

Class 9
Subject Maths
Book Mathematics
Chapter Number 14
Chapter Name  

Statistics

NCERT Solutions Class 9 Maths chapter 14 Statistics

Class 9, Maths chapter 14, Statistics solutions are given below in PDF format. You can view them online or download PDF file for future use.

Statistics

Q.1: Give five examples of data that you can collect from your day-to-day life.

Ans : In our day to life, we can collect the following data. 1.Number Of females per 1000 males in various States Of our country. 2. Weights of students of our class . 3. Production of wheat in the last 10 years in our country . 4. Number of plants in our locality . 5. Rainfall in our city in the last 10 years

Q.2: Classify the data in Q.1 above as primary or secondary data.

Ans : The information which is collected by the investigator with a definite objective in his mind is called as primary data whereas when the information is gathered from a source which already had the information stored, it is called as secondary data. It can be observed that the data in 1, 3, and 5 IS secondary data and the data in 2 and 4 is primary data.

Q.1: The blood groups of 30 students of Class VIII are recorded as follows:
 A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
 A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?

Ans : It can be observed that 9 students have their blood group as A, 6 as B, 3 as Ad, and 12 as O. Therefore, the blood group of 30 students of the class can be represented as follows. It can be observed clearly that the most common blood group and the rarest blood group among these students is O and AB respectively as 12 (maximum number of students) have their blood group as O, and 3 (minimum number of students) have their blood group as AB.

Q.2: The distance (in km) of 40 engineers from their residence to their place of work were found as follows: 
5   3    10  20  25  11  13  7   12  31
19 10  12  17  18  11  32  16  2     7
7   9    7    8    3    5    12  15 18   3 
12 14  2    9    6    15  15  7    6   12 
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?

Ans : Let us make the classes as 0-5, 5-10, 10-15, 15-20, 20-25, 25-30 and 30-35. The frequency distribution table will be as follows:

Q.3: The relative humidity (in %) of a certain city for a month of 30 days was as follows:
 98.1 98.6 99.2 90.3 86.5 95.3 92.9 96.3 94.2 95.1 
89.2 92.3 97.1 93.5 92.7 95.1 97.2 93.3 95.2 97.3 
96.2 92.1 84.9 90.2 95.7 98.3 97.3 96.1 92.1 89 
(i) Construct a grouped frequency distribution table with classes 84 - 86, 86 - 88, etc. 
(ii) Which month or season do you think this data is about? 
(iii) What is the range of this data?

Ans : (i) A grouped frequency distribution table of class size 2 has to be constructed. The class intervals will be -84 - 86, 86 - 88, and -99 - 90... By observing the data given above, the required table can be constructed as follows. (ii) It can be observed that the relative humidity is high. Therefore, the data is about a month of rainy season. (iii) Range of data = Maximum value — Minimum value =- 99.2 + 84.9 =- 14.3

Q.4: The heights of 50 students, measured to the nearest centimetres, have been found to be as follows: 
161 150 154 165 168 161 154 162 150 151 
162 164 171 165 158 154 156 172 160 170 
153 159 161 170 162 165 166 168 165 164 
154 152 153 156 158 162 160 161 173 166 
161 159 162 167 168 159 158 153 154 159 
(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 - 165, 165 - 170, etc.
(ii) What can you conclude about their heights from the table?

Ans : (i) A grouped frequency distribution table has to be constructed taking class intervals 160 -165, 165 - 170, etc. dy observing the data given above, the required table can be constructed as follows. (ii) It can be concluded that more than 50% of the students are shorter than 165 .

Q.5: A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:
 0.03  0.08  0.08   0.09   0.04   0.17
 0.16  0.05  0.02   0.06   0.18  0.20 
 0.11   0.08  0.12  0.13  0.22   0.07 
 0.08   0.01   0.10  0.06  0.09   0.18    
 0.11   0.07   0.05   0.07   0.01     0.04 
(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 - 0.04, 0.04 - 0.08, and so on. 
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?

Ans : Taking class intervals as 0.00, —0.04, 0.04, —0.08, and so on, a grouped frequency table can be constructed as follows. The number of days for which the concentration of \( \mathrm{SO}_{2}\) is more than 0.11 is the number of days for concentration is in between 0.12 - 0.16, 0.16 - 0.20, 0.20 - 0.24 . Required number of days=2+4+2=8 There,for 8 days, the concentration of \( \mathrm{SO}_{2}\) is more than 0.11 ppm.

Q.6: Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows: 
0  1  2  2  1  2  3  1  3  0 
1  3  1  1  2  2  0  1  2 1 
3 0  0  1  1  2  3  2  2  0 
Prepare a frequency distribution table for the data given above.

Ans : By the observation the data given above, the required frequency distribution table can be constructed as follows.

Q.7: The value of π upto 50 decimal places is given below: 
3.14159265358979323846264338327950288419716939937510 
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point. 
(ii) What are the most and the least frequently occurring digits?

Ans : (i) By the observation of the digits after decimal point, the required table can be constructed as follows. (ii) it can observed from the table that the least frequency is 2 of digit 0, and the maximum frequency is 8 of digit 3 and 9 .therefor , the most frequently occurring digits are 3 and 9 and the least frequency occurring digit is 0.

Q.8: Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows: 
1    6  2  3  5  12  5  8  4 8 
10  3  4  12  2 8 15 1 17  6 
3   2  8  5  9 6  8 7  14  12 
(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5 - 10. 
(ii) How many children watched television for 15 or more hours a week?

Ans : (i) our class interval will be 0 - 5 , 5 - 10 , 10 - 15 …… The group distribution frequency table can be constructed as follows. (ii) the number of children who watched TV for 15 or more hours a week is 2 (i.e…,the number of children in class interval 15-20 )

Q.9: A company manufactures car batteries of particular type. The lives of 4 sch batteries were recorded as fellows.
2.6  3.0  3.7  3.2  2.2  4.1  3.5  4.5 
3.5  2.3  3.2  3.4  3.8  3.2  4.6  3.7 
2.5  4.4  3.4  3.3  2.9  3.0  4.3  2.8 
3.5  3.2  3.9  3.2  3.2  3.1  3.7  3.4 
4.6  3.8  3.2  2.6  3.5  4.2  2.9  3.6 
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 - 2.5.

Ans : A grouped frequency table of class size 0.5 has to be constructed, starting from class interval 2 - 2.5. Therefore, the class intervals Will be 2 - 2.5, 2.5 - 3, 3 - 3.5... By observing the data given above, the required grouped frequency distribution table can be constructed as follows.

Q.1: A survey conducted by an organisation for the cause of illness and death among the women between the ages 15 - 44 (in years) worldwide, found the following figures (in %):



(i) Represent the information given above graphically. 
(ii) Which condition is the major cause of women's ill health and death worldwide? 
(iii) Try to find out, With the help Of your teacher, anv two factors which play a major role in the cause in (ii) above being the major cause.

Ans : (i) By representing causes on x-axis and family fatality rate on y-axis and choosing an appropriate scale (I unit 5% for y axis), the graph of the information given above can be constructed as follows. All the rectangle bars are Of the same width and have equal spacing between them. (ii) Reproductive health condition is the major cause of women's ill health and death worldwide as 31.8% of women are affected it. (iii) The factors are as follows. 1. Lack of medical facilities 2. Lack of correct knowledge of treatment

Q.2: The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.

(i) Represent the information above a bar graph .
(ii) In the classroom discuss what conclusions can be arrived at from the graph.

Ans : (i) By representing section (variable) on x-axis and number of girls per thousand boys on y-axis, the graph of the information given above can be constructed by choosing an appropriate scale (1 unit 100 girls for paxis) Here, all the rectangle bars are of the same length and have equal spacing in between them. (ii) It can be observed that maximum number Of girls per thousand boys (i.e., 970) is for ST and minimum number Of girls per thousand boys (i.e., 910) is for urban. Also, the number of girls per thousand boys is greater in rural areas than that in urban areas, backward districts than that in non-backward districts, SC and ST than that in non-SC/ST.

Q.3: Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

(i) draw a bar graph to represent the polling results.
(ii) which political party won the maximum number of seats.

Ans : (i) By taking polling results on X-axis and seats won as y-axis and choosing an appropriate scale (1 unit -=10 seats for y-axis), the required graph of the above information can be constructed as follows. Here, the rectangle bars are of the same length and have equal spacing in between them. (ii) Political party 'A' won maximum number of seats.

Q.4: The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:

(i) Draw a histogram to represent the given data. 
(ii) Is there any other suitable graphical representation for the same data? 
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?

Ans : (i) It can be observed that the length of leaves is represented in a discontinuous class interval having a difference of 1 in between them. Therefore, \( \frac{1}{2}=0.5\) has to be added to each upper class limit and also have to subtract 0.5 from the class limits so as to make the class intervals continuous. Taking the length of leaves on x-axis and the number of leaves on y-axis, histogram of this information can be drawn as above. Here, 1 unit on y-axis represents 2 leaves. (ii) Other suitable graphical representation of this data is frequency polygon. (iii) NO, as maximum number Of leaves (i.e., 12) has their length in between 144.5 mm and 153.5 mm. It is not necessary that all ha'.oe their lengths as 153 mm.

Q.5: The following table gives the life times of 400 neon lamps:

(i) Represent the given information with the help of a histogram. 
(ii) How many lamps have a life-time of more than 700 hours?

Ans : (i) By taking life time (in hours) of neon lamps on x-axis and the number of lamps on y-axis, the histogram of the given information can be drawn as follows. \ Here, 1 unit on y-axis represents 10 lamps. (ii) It can be concluded that the number of neon lamps having their lifetime more than 700 is the sum of the number of neon lamps having their lifetime as 700 — 800, 800 - goo, and 900 - 1000. Therefore, the number of neon lamps having their lifetime more than 700 hours is 184. (74 + 62 +48 - 184)

Q.6: The following table gives the distribution of students of two sections according to the marks obtained by them:
 

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.

Ans : We can find the class marks Of the given class intervals by using the following formula. Class mark =\( \frac{\text { Upper class limit }+\text { Lower class limit }}{2}\) Taking class marks on x-axis and frequency on y-axis and choosing an appropriate scale (1 unit - 3 for y-axis), the frequency polygon can be drawn as follows. It can be observed that the performance Of students Of section 'A' is better than the students of section 'B' in terms of good marks.

Q.7: The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:
 

Represent the data of both the teams on the same graph by frequency polygons.

Ans : It can be observed that the class inter,zals of the given data are not continuous. There is a gap of I in between them. Therefore, \( \frac{1}{2}=0.5\) has to be added to the upper class limits and 0.5 has to be subtracted from the lower class limits. Also, class mark of each interval can be found by using the following formula. Class mark \( =\frac{\text { Upper class limit }+\text { Lower class limit }}{2}\) Continuous data with class mark of each class interval can Be represented as follows. By taking class marks on X-axis and runs scored on y-axis, a frequency polygon can be constructed as follows.

Q.8: A random survey of the number of children of various age groups playing in a park was found as follows:

Draw a histogram to represent the data above.

Ans : Here, it can be observed that the data has class intervals of varying width. The proportion Of children per 1 year interval can be calculated as follows. Taking the age of children on x-axis and proportion children per 1 year interval on y-axis, the histogram can be drawn as follows.

Q.9: 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:


(i) Draw a histogram to depict the given information. 
(ii) Write the class interval in which the maximum number of surnames lie.

Ans : (i) Here, it can be observed that the data has class intervals of varying width. The proportion of the number of surnames per 2 letters interval can be calculated as follows. By taking the number of letters on X-axis and the proportion of the number of surnames per 2 letters interval on y-axis and choosing an appropriate scale (1 unit - 4 students for y axis), the histogram can be constructed as follows. (ii) The class interval in which the maximum number of surnames lies is 6 - 8 as ithas 44 surnames in it i.e., the maximum for this data.

Q.1: The following number of goals were scored by a team in a series of 10 matches: 
2, 3, 4, 5, 0, 1, 3, 3, 4, 3

Ans : The number of goals scored by the team is 2,3,4,5,0,1,3,3,4,3 Mean of data \( =\frac{\text { Sum of all observations }}{\text { Total number of observations }}\) Mean score \( =\frac{2+3+4+5+0+1+3+3+4+3}{10}\) \( =\frac{28}{10}=2.8\) =2.8 goals Arranging the number of goals in ascending order, 0,1,2,3,3,3,3,4,4,5 The number Of observations is 10, which is an even number. Therefore, median Score will be the mean of \( \frac{10}{2}\) i.e.,5th and \( \frac{10}{2}+1\) i.e,6th observation while arranged in ascending or descending order Median score= \( \frac{5^{\text { th }} \text { observation }+6^{\text { th }} \text { observation }}{2}\) \( =\frac{3+3}{2}\) \( =\frac{6}{2}\) =3 Mode Of data is the observation With the maximum frequency in data. Therefore, the mode score of data is 3 as has the maximum frequency as 4 in the data.

Q.2: In a mathematics test given to 15 students, the following marks (out of 100) are recorded: 
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60  Find the mean, median and mode of this data.

Ans : The marks of 15 students in mathematics test are 41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60 Mean of data \( =\frac{\text { Sum of all observation }}{\text { Total number of observation }}\) \( \frac{41+39+48+52+46+62+54+40+96+52+40+40+42+60}{15}\) \( =\frac{822}{15}=54.8\) Arranging the Scores obtained by 15 students in an ascending order, 39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, gs As the number Of observations is 15 which is Odd, therefore, the median Of data Will \( \frac{15+1}{2}\)= 8th observation whether the data is arranged in an ascending or descending order. Therefore, median score of data = 52 Made of data is observation with the maximum frequency in data. Therefore mode of this data is 52 having the highest frequency in data a 3.

Q.3: The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95

Ans : It can be observed that the total number of observations in the given data is 10 (even number) . therefore the median of this data will be the mean of \( \frac{10}{2}\),5th and \( \frac{10}{2}+1\) i.e.,6th observation. Therefore, median of data = \( \frac{5^{\text { th }} \text { observation }+6^{\text { th }} \text { observation }}{2}\) \( 63=\frac{x+x+2}{2}\) \( 63=\frac{2 x+2}{2}\) \( 63=x+1\) \( x=62\)

Q.4: . Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.

Ans : Arranging the data in an ascending order, 14, 14, 14, 14, 17, 18, IS, 18, 22, 23, 25, 28 It can be observed that 14 has the highest frequency, i.e. 4, in the given data. Therefore, mode of the given data is 14.

Q.5: Find the mean salary of 60 workers of a factory from the following table:

Ans : We know that Mean=\( \frac{\sum f_{i} x_{i}}{\sum f_{i}}\) The value of \( \sum f_{i} x_{i}\) and \( \sum f_{i}\) can be calculated as follows: Mean salary= \( \frac{305000}{60}\) =5083.33 Therefore,mean salary of 60 workers is Rs. 5083.33

Q.6: Give one example of a situation in which  
(i) the mean is an appropriate measure of central tendency. 
(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.

Ans : When any data has a few observations such that these are very far from the other observations in it, it is better to calculate the median than the mean of the data as median gives a better estimate of average in this case. (i) Consider the following example — the following data represents the heights of the members of a family. 154.9 cm, 162.8 cm, 170.6 cm, 158.8 cm, 163.3 cm, 166.8 cm, 160.2 cm In this case, it can be observed that the observations in the given data are close to each other. Therefore, mean will be calculated as an appropriate measure of central tendency. (ii) The following data represents the marks obtained by 12 students in a test. 48, 59, 46, 52, 54, 46, 97, 42, 49, 58, 60, gg In this case, it can be observed that there are some observations which are very far from other observations. Therefore, here, median will be calculated as an appropriate measure of central tendency.

Q.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.

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

Ans : To find the class mark \( \left(x_{i}\right) \) for each interval, the following relation is used. Class mark \( \left(x_{i}\right) \) \( \frac{\text { Upper class limit + Lower class limit }}{2} \) \( x_{i} \text { and } f_{X_{i}} \) can be calculated as follows From the table it can be observed that \( \sum f_{i}=20 \) \( \sum f_{r} x_{i}=162 \) Mean \( \overline{x}=\frac{\sum f_{f_{i}}}{\sum f_{i}} \) \( =\frac{162}{20}=8.1 \) Therefore mean number of plants per house is 8.1 Here, direct method has been used as the values of class marks \( \left(x_{i}\right) \) and \( f_{i} \) are small

Q.2: Consider the following distribution of daily wages of 50 workers of a factory

Find the mean daily wages of the workers of the factory by using an appropriate method

Ans : To find the class mark for each interval, the following relation is used, \( x_{i}=\frac{\text { Upper class limit }+\text { Lower class limit }}{2} \) Class size (h ) of this data =20 Taking 150 as assured mean (a) d, u and f can be calculated as follows. From the table it can be observed that \( \sum f_{i}=50 \) \( \sum f_{i} u_{i}=-12 \) Mean \( \overline{x}=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) h \) \( =150+\left(\frac{-12}{50}\right) 20 \) \( =150-\frac{24}{5} \) \(=150-4.8 \) \( =145.2 \) Therefore the mean daily wage of the workers of the factory is Rs 145.20

Q.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.

Ans : To find the class mark \(\left(x_{i}\right) \) for each interval, the following relation is used. \( x_{i}=\frac{\text { Upper class limit }+\text { Lower class limit }}{2} \) Given that mean pocket allowance \( \overline{x}=\operatorname{Rs} 18 \) Taking 18 as assured mean \( (a), d_{i} \text { and } f d_{i} \) are calculated as follows. From the table, we obtain \( \sum f_{i}=44+f \) \( \sum f_{i} u_{l}=2 f-40 \) \(\overline{x}=a+\frac{\sum f_{i} d_{i}}{\sum f_{l}} \) \( 18=18+\left(\frac{2 f-40}{44+f}\right) \) \( 0=\left(\frac{2 f-40}{44+f}\right) \) \( 2 f-40=0 \) \( 2 f=40 \) \( f=20 \) Hence the missing frequency f is 20

Q.4: Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Ans : To find the class mark of each interval (x,), the following relation is used. \( x_{i}=\frac{\text { Upper class limit }+\text { Lower class limit }}{2} \) Class size h of this data =3 Taking 75.5 as assumed mean \( (a), d i, u_{i t}, f_{i} u_{i} \) are calculated as follows. From the table we obtain \( \sum f i=30 \) \(\sum f_{\mu,}=4 \) Mean \( \overline{x}=a+\left(\frac{\sum f u_{i}}{\sum f_{i}}\right) \times h \) \(=75.5+\left(\frac{4}{30}\right) \times 3 \) \( =75.5+0.4=75.9 \) Therefore ,mean hear beats per minute for these women are 75.9 beats per minute.

Q.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.

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

Ans : It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore, 1/2 has to be added to the upper class limit and 1/2 has to be subtracted from the lower class limit of each interval. Class mark (x,) can be obtained by using the following elation. \( x_{i}=\frac{\text { Upper class limit + Lower class limit }}{2} \) Class size (h) of this data = 3 Taking 57 as assumed mean , \((a), d_{i}, u_{i t} f_{\mu_{i}} \) are calculated as follows. It can be observed that \( \sum f_{i}=400 \) \(\sum f_{i} u_{i}=25 \) Mean \( \overline{x}=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) \times h \) \(=57+\left(\frac{25}{400}\right) \times 3 \) \( =57+\frac{3}{16}=57+0.1875 \) \( =57.1875 \) \( =57.19 \) Mean number of mangoes kept in a packing box is 57.19 Step deviation method is used here as the values of \( f_{i,} d_{i} \) are big and also there is a common multiple between all \( d_{i} \)

Q.6: The table below shows the daily expenditure on food of 25 households in a locality.
Find the mean daily expenditure on food by a suitable method.

Ans : To find the class mark \(\left(X_{i}\right) \) for each interval, the following relation is used, \( x_{1}=\frac{\text { Upper class limit }+\text { Lower class limit }}{2} \) Class size =50 Taking 225 as assumed mean \( (a), d_{i}, u_{i}, f_{\mu_{i}} \) are calculated as follows. From the table , we obtain \( \Sigma f_{i}=25 \) \( \sum f_{i} u_{i}=-7 \) Mean \( \overline{x}=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) \times h \) \( =225+\left(\frac{-7}{25}\right) \times(50) \) \( =225-14 \) \( =211 \) Therefore, mean daily expenditure on food is Rs 211

Q.7: To find out the concentration of \(  \mathrm{SO}_{2} \) 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: 

Find the mean concentration of \(  \mathrm{SO}_{2} \)  in the air.

Ans : To find the class marks for each interval the following relation is used \(x_{i}=\frac{\text { Upper class limit }+\text { Lower class limit }}{2} \) Class sizeof this data =0.04 Taking 0.14 as assumed mean \( (a), d_{i t} u_{i r} f_{\mu_{i}} \) are calculated as follows. From the table we obtain \( \sum f_{i}=30 \) \( \sum f_{i} u_{i}=-31 \) Mean \( \overline{x}=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) \times h \) \( =0.14+\left(\frac{-31}{30}\right)(0.04) \) \( =0.14-0.04133 \) \( =0.09867 \) \(=0.099 \mathrm{ppm} \) Therefore mean concentration of \( \mathrm{SO}_{2} \) in the air is 0.099 ppm

Q.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

Ans : To find the class mark of each interval the following relation is used \( x_{i}=\frac{\text { Upper class limit + Lower class limit }}{2} \) Taking 17 as assumed mean \( (a), d_{i} \text { and } f_{d_{i}} \)are calculated as follows. From the table we obtain \( \Sigma f_{i}=40 \) \(\Sigma f_{i} d_{i}=-181 \) Mean \( \overline{x}=a+\left(\frac{\sum f_{j} d_{i}}{\sum f_{i}}\right) \) \( =17+\left(\frac{-181}{40}\right) \) \( =17-4.525 \) \( =12.475 \) \( =12.48 \) Therefore the mean number of days is 12.48 for which a student was absent.

Q.9: The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Ans : To find the class marks the following relation is used \( x_{i}=\frac{\text { Upper class limit }+\text { Lower class limit }}{2} \) Class size (h) for this data =10 Taking 70 as assumed mean \( (a), d_{i}, u_{i t} \text { and } f_{\prime} u_{i} \) are calculated as follows From the table we obtain \( \Sigma f_{i}=35 \) \( \sum f_{i} u_{i}=-2 \) Mean \( \overline{x}=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{f}}\right) \times h \) \( =70+\left(\frac{-2}{35}\right) \times(10) \) \( =70-\frac{20}{35} \) \( =70-\frac{4}{7} \) \( =70-0.57 \) \( =69.43 \) Therefore ,mean literacy rate is 69.43%

Q.1: The following table shows the ages of the patients admitted in a hospital during a year 

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

Ans : To find the class marks \( \left(x_{i}\right) \) the following relation is used \(x_{i}=\frac{\text { Upper class limit + Lower class limit }}{2} \) Taking 30 as assumed mean \(d_{i} \text { and } f_{f} d_{i} \) are calculated as follows. From the table we obtain \( \begin{aligned} \sum f_{i}=80 & \\ \sum f_{i} d_{i}=& 430 \\ \text { Mean, } \overline{x} &=a+\frac{\sum f_{l} d_{j}}{\sum f_{i}} \\ &=30+\left(\frac{430}{80}\right) \\ &=30+5.375 \\ &=35.375 \\ &=35.38 \end{aligned} \) Mean of this data 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 \( \left(f_{1}\right)\) of modal class =23 Class size (h) =10 Frequency \( \left(f_{0}\right) \) of class preceding the modal class =21 Frequency \( \left(f_{2}\right) \) of class succeeding the modal class =14 Mode= \( \begin{aligned} & l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h \\ &=35+\left(\frac{23-21}{2(23)-21-14}\right) \times 10 \\ &=35+\left[\frac{2}{46-35}\right] \times 10 \\ &=35+\frac{20}{11} \\ &=35+1.81 \\ &=36.8 \end{aligned} \) Mode is 36.8 It represents that the age of maximum number of patients admitted in hospital was 36.8 years.

Q.2: The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Ans : 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 \( \left(f_{1}\right) \) of modal class = 61 Frequency \(\left(f_{0}\right) \) of class preceding the modal class = 52 Frequency \( \left(f_{2}\right) \) of class succeeding the modal class 38 Class size (h) = 20 \( \begin{aligned} \text { Mode } &=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h \\ &=60+\left(\frac{61-52}{2(61)-52-38}\right)(20) \end{aligned} \) \( \begin{array}{l}{=60+\left(\frac{9}{122-90}\right)(20)} \\ {=60+\left(\frac{9 \times 20}{32}\right)} \\ {=60+\frac{90}{16}=60+5.625} \\ {=65.625}\end{array} \) Therefore ,modal lifetime of electrical components is 65.625 hours.

Q.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. 

Ans : It can be observed from the given data that the maximum class frequency s 40, belonging to 1500 — 2000 inte vals. Therefore, modal class =1500 — 2000 Lower limit (l) of modal class = 1500 Frequency \( \left(f_{1}\right) \)of modal class = 40 Frequency \( \left(f_{0}\right) \) of class preceding modal class=24 Frequency \( \left(f_{2}\right) \) of class succeeding modal class = 33 Class size (h) =500 \( \begin{aligned} \text { Mode } &=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h \\ &=1500+\left(\frac{40-24}{2(40)-24-33}\right) \times 500 \\ &=1500+\left(\frac{16}{80-57}\right) \times 500 \\ &=1500+\frac{8000}{23} \\ &=1847.826=1847.83 \end{aligned} \) Therefore , modal monthly expenditure was Rs 1847.83 To find the class mark , the following relation is used Class mark \( =\frac{\text { Upper class limit }+\text { Lower class limit }}{2} \) Class Size(h) of the given data=500 Talking 2750 as assumed mean \( (a), d_{i}, u_{i}, \text { and } f_{\mu}\) are calculated as follows. From the table we obtain \(\begin{array}{l}{\sum f_{i}=200} \\ {\sum f_{i} u_{f}=-35} \\ {\overline{x}(\text { mean })=a+\left(\frac{\sum f_{i} u_{l}}{\sum f_{i}}\right) \times h} \\ {\overline{x}=2750+\left(\frac{-35}{200}\right) \times 500} \\ {=2750-87.5} \\ {=2662.5}\end{array} \) Therefore, mean monthly expenditure was Rs 2662.50

Q.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.

Ans : It can be observed from the given data that the maximum class frequency s 10 belonging to class interval 30 — 35. Therefore, modal class = 30 — 35 Class size (h) = 5 Lower limit (l) of modal class =30 Frequency \( \left(f_{1}\right) \) of modal class = 10 Frequency \(\left(f_{0}\right) \) of class preceding modal class = 9 Frequency \( \left(f_{2}\right) \) of class succeeding modal class = 3 \( \begin{aligned} \operatorname{Mode} &=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h \\ &=30+\left(\frac{10-9}{2(10)-9-3}\right) \times(5) \\ &=30+\left(\frac{1}{20-12}\right)^{5} \\ &=30+\frac{5}{8}=30.625 \\ \text { Mode } &=30.6 \end{aligned} \) It represents that most of the states/LJ.T have a teacher-student ratio as 30.6. To find the class marks, the following relation is used. Class mark \(=\frac{\text { Upper class limit }+\text { Lower class limit }}{2} \) Taking 32.5 as assumed mean \( d_{i}, u_{i}, \text { and } f_{\prime} \mu_{i} \) are calculated as follows. \( \begin{aligned} \text { Mean, } \overline{x} &=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) h \\ &=32.5+\left(\frac{-23}{35}\right) \times 5 \\ &=32.5-\frac{23}{7}=32.5-3.28 \\ &=29.22 \end{aligned} \) Therefore ,mean of the data is 29.2 It represents that on an average , teacher-student ratio was 29.2

Q.5: The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data

Ans : 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 \( \left(f_{1}\right) \) of modal class 18 Frequency \( \left(f_{0}\right) \) of class preceding modal class =4 Frequency \( \left(f_{2}\right) \) of class succeeding modal class = 9 Class size = 1000 \( \begin{aligned} \operatorname{Mod} e &=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h \\ &=4000+\left(\frac{18-4}{2(18)-4-9}\right) \times 1000 \\ &=4000+\left(\frac{14000}{23}\right) \\ &=4000+608.695 \\ &=4608.695 \end{aligned} \) Therefore mode of the given data is 4608.7 runs

Q.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 table given below. Find the mode of the data:

Ans : 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 \( \)of modal class = 40 Frequency \( \left(f_{1}\right) \) of modal class =20 Frequency \( \left(f_{0}\right) \) of class preceding modal class=12 Frequency \( \left(f_{2}\right) \) of class succeeding modal class =11 Class size = 10 \( \begin{aligned} \text { Mode } &=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h \\ &=40+\left[\frac{20-12}{2(20)-12-11}\right] \times 10 \\ &=40+\left(\frac{80}{40-23}\right) \\=& 40+\frac{80}{17} \\=& 40+4.7 \\=& 44.7 \end{aligned} \) Therefore , mode of this data is 44.7 cars

Q.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

Ans : To find the class marks the following relation is used \( =\frac{\text { Upper class limit + }+\text { Lower class limit }}{2} \) Taking 135 as assumed mean \( (a), d_{i} u_{i n} f \mu_{i} \) are calculated according to step deviation method as follows From the table ,we obtain \( \begin{array}{l}{\sum f_{i} u_{i}=7} \\ {\sum f_{i}=68} \\ {\text { Class size }(h)=20}\end{array} \) \( \begin{aligned} \text { Mean, } \overline{x} &=a+\left(\frac{\sum f u_{i}}{\sum f_{i}}\right) \times h \\ &=135+\frac{7}{68} \times 20 \\ &=135+\frac{140}{68} \\ &=137.058 \end{aligned} \) 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 \( \left(f_{1}\right) \) of modal class =20 Frequency \( \left(f_{0}\right) \) of class preceding modal class = 13 Frequency\( \left(f_{2}\right) \) of class succeeding the modal class = 14 \( \begin{aligned} \text { Mode } &=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h \\ &=125+\left[\frac{20-13}{2(20)-13-14}\right] \times 20 \\ &=125+\frac{7}{13} \times 20 \\ &=125+\frac{140}{13}=135.76 \end{aligned} \) To find the median of the given data, cumulative frequency is calculated as follows. From the table we obtain n=68 Cumulative frequency (cf) just greater than interval 125 - 145. Therefore, median class = 125 — 145 Lower limit (I) of median class =125 Class size (h) =20 Frequency (cf) of median class =20 Cumulative frequency (cf) of class preceding median class = 22 \(\begin{aligned} \text { Median } &=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \\ &=125+\left(\frac{34-22}{20}\right) \times 20 \\ &=125+12 \\ &=137 \end{aligned} \) Therefore median mode mean of the given data is 137, 135.76 and 137.05 respectively. The three measures are approximately the same .

Q.2: If the median of the distribution given below is 28.5, find the values of x and y.

Ans : The cumulative frequency for Hoe given data is calculated as follows. From the table it can be observed that n=60 \( \begin{array}{l}{45+x+y=60} \\ {x+y=15(1)}\end{array} \) 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 \(\begin{array}{l}{\text { Median }=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h} \\ {28.5=20+\left[\frac{\frac{60}{2}-(5+x)}{20}\right] \times 10} \\ {8.5=\left(\frac{25-x}{2}\right)} \\ {17=25-x} \\ {x=8}\end{array}\) From equation (1) \( \begin{array}{l}{8+y=15} \\ {y=7}\end{array} \) Hence the value of x and y are 8 and 7 respectively

Q.3: 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.

Ans : 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 From the table, it can be observed that n = 100 Cumulative frequency (cf) just greater than \(\frac{n}{2}\left(\text { i.e., } \frac{100}{2}=50\right) \) 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 \( \begin{aligned} \text { Median } &=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \\ &=35+\left(\frac{50-45}{33}\right) \times 5 \\ &=35+\frac{25}{33} \\ &=35.76 \end{aligned} \) Therefore ,median age is 35.76 years.

Q.4: 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: 

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)

Ans : The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore, \( \frac{1}{2}=0.5 \) 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. From the table it can be observed that the cumulative frequency just greater than \( \frac{n}{2}\left(\text { i.e., } \frac{40}{2}=20\right) \) 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 Median \( =l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \) \( \begin{array}{l}{=144.5+\left(\frac{20-17}{12}\right) \times 9} \\ {=144.5+\frac{9}{4}=146.75}\end{array} \) Therefore median length of leaves is 146.75 mm

Q.5: The following table gives the distribution of the life time of 400 neon lamps :

Find the median lifetime of a lamp.

Ans : The cumulative frequencies with their respective class intervals are as follows. It can be observed that the cumulative frequency just greater than \( \frac{n}{2}\left(\text { i.e., } \frac{400}{2}=200\right) \) 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 Median \( =l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \) \( \begin{aligned} &=3000+\left(\frac{200-130}{86}\right) \times 500 \\ &=3000+\frac{70 \times 500}{86} \\ &=3406.976 \end{aligned} \) Therefore, median lifetime of lamps is 3406.98 hours.

Q.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:

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.

Ans : The cumulative frequencies with their respective class intervals are as follows. It can be observed that the cumulative frequency just greater than \( \frac{n}{2}\left(\text { i.e., } \frac{100}{2}=50\right) \) is 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 Median \( =l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \) \( \begin{aligned} &=7+\left(\frac{50-36}{40}\right) \times 3 \\ &=7+\frac{14 \times 3}{40} \\ &=8.05 \end{aligned} \) To find the class marks of the given class intervals the following relation is used Class mark \(=\frac{\text { Upper class limit }+\text { Lower class limit }}{2} \) Taking 11.5 as assumed mean \( d_{i t} u_{i r} \text { and } f_{,} u_{i} \) are calculated according to step deviation method as follows . From the given table \( \begin{array}{l}{\sum f_{i} \mu_{i}=-106} \\ {\Sigma f_{i}=100}\end{array} \) \( \begin{array}{l}{\overline{x}=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) h} \\ {=11.5+\left(\frac{-106}{100}\right) \times 3} \\ {=11.5-3.18=8.32}\end{array} \) The data in the given table can be written as From the table, it can be observed that the maximum class frequency is 40 belonging to class interval 7 — 10 Modal class =7 — 10 Lower limit (l) of modal class = 7 Class size (h) = 3 Frequency \( \left(f_{1}\right) \) of modal class == 40 Frequency \(\left(f_{0}\right) \) Of class preceding the modal class = 30 Frequency \( \left(f_{2}\right) \) of class succeeding the modal class =16 \(\begin{aligned} \text { Mode }=& l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h \\ &=7+\left[\frac{40-30}{2(40)-30-16}\right] \times 3 \\ &=7+\frac{10}{34} \times 3 \\ &=7+\frac{30}{34}=7.88 \end{aligned} \) Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.88.

Q.7: The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Ans : The cumulative frequencies with their respective class intervals are as follows. Cumulative frequency just greater than interval 55 — 60. Median class = 55 — 60 Lower limit (l) of median class=55 Frequency (f) of median class = 6 is 19, belonging to class Cumulative frequency (cf) of median class = 13 Class size (h) = 5 Median \(=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \) \( \begin{array}{l}{=55+\left(\frac{15-13}{6}\right) \times 5} \\ {=55+\frac{10}{6}} \\ {=56.67}\end{array} \) Therefore the median weight is 56.67 kg

Q.1: The following distribution gives the daily income of 50 workers of a factory.

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

Ans : The frequency distribution table of less than type is as follows. Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows.

Q.2: During the medical check-up of 35 students of a class, their weights were recorded as follows:

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.

Ans : The given cumulative frequency distributions of less than type are 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 , \( \frac{n}{2}=17.5 \) Mark the point A whose ordinate is 17.5 and its co-ordinate is 46.5 .Therefore median of this data is 46.5 It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below. The cumulative frequency just greater than \( \frac{n}{2}\left(\text { i.e., } \frac{35}{2}=17.5\right) \) 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 \( \begin{aligned} \text { Median } &=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \\ &=46+\left(\frac{17.5-14}{14}\right) \times 2 \\ &=46+\frac{3.5}{7} \\ &=46.5 \end{aligned} \) Therefore, median of this data is 46.5. Hence, the value of median is verified.

Q.3: The following table gives production yield per hectare of wheat of 100 farms of a village.

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

Ans : The cumulative frequency distribution of more than type can be obtained as follows. Taking the lower class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be obtained as follows.

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