# Statistics

## Class 10 NCERT

### NCERT

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?

##### Solution :

To find the class mark $(x_i)$ for each interval, the following relation is used.$\\$ $X_i=\dfrac{Upper \ class \ limit \ + \ Lower \ class \ limit}{2}$$\\ Given that, mean pocket allowance,\bar{X} =Rs 18$$\\$ Taking $18$ as assured mean $(a), d _i$ and $f _i d _i$ are calculated as follows:$\\$ table $\\$ From the table, we obtain$\\$ $\Sigma f_i=44+f\\ \Sigma f_i u_i=2f-40\\ \bar{x} =a+\dfrac{\Sigma f_id_i}{\Sigma f_i}\\ 18= 18+(\dfrac{2f-40}{44+f})\\ 2f-40=0\\ 2f=40\\ f=20$Hence, the missing frequency $f_ i$ is $20$.

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

##### Solution :

To find the class mark of each interval $(x_i)$, the following relation is used.$\\$ $X_i=\dfrac{Upper \ class \ limit \ + \ Lower \ class \ limit}{2}$$\\ Class size, h, of this data = 3 Taking 75.5 as assumed mean (a), d_ i , u _i , f_ i u _i are calculated as follows:\\ table \\ From the table, we obtain\\ Mean \bar{X} =a+(\dfrac{\Sigma f_i u_i}{\Sigma f_i})*h\\ =75.5+(\dfrac{4}{30})*3\\ =75.5 +0.4 =75.9$$\\$ Therefore, mean hear beats per minute for these women are $75.9$ beats per minute.

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?

##### Solution :

It can be observed that class intervals are not continuous. There is a gap of $1$ between two class intervals. Therefore,$\dfrac{1}{2}$ has to be added to the upper class limit and $\dfrac{1}{2}$ has to be subtracted from the lower class limit of each interval. Class mark (xi) can be obtained by using the following relation.$\\$ $X_i=\dfrac{Upper \ class \ limit \ + \ Lower \ class \ limit}{2}$$\\Class size (h) of this data = 3$$\\$ Taking $57$ as assumed mean $(a), d_ i , u_ i , f_ i u_ i$ are calculated as follows.$\\$ table $\\$ It can be observed that$\\$ $\Sigma f_i =400\\ \Sigma f_i u_i =25$$\\ Mean \bar{x} =a+(\dfrac{\Sigma f_i u_i}{\Sigma f_i})*h\\ =57+(\dfrac{25}{400})*3\\ = 57+\dfrac{3}{16}=57+0.1875\\ =57.19$$\\$ 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 $d_ i$ .

6   The table below shows the daily expenditure on food of $254$ households in a locality.$\\$ table $\\$ Find the mean daily expenditure on food by a suitable method.

To find the class mark (xi) for each interval, the following relation is used.$\\$ $X_i=\dfrac{Upper \ class \ limit \ + \ Lower \ class \ limit}{2}$$\\ Class size = 50 Taking 225 as assumed mean (a), d_i, u_i, f_iu_i are calculated as follows:\\ table \\ From the table we obtain\\ \Sigma f_i =25\\ \Sigma f_i u_i=-7$$\\$ Mean $\bar{x} =a+(\dfrac{\Sigma f_i u_i}{\Sigma f_i})*h\\ =225+(\dfrac{-7}{25})*(50)\\ =211$$\\ Therefore, mean daily expenditure on food is Rs 211. 7 To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:\\ table \\ Find the mean concentration of SO_ 2 in the air. ##### Solution : To find the class marks for each interval, the following relation is used.\\ X_i=\dfrac{Upper \ class \ limit \ + \ Lower \ class \ limit}{2}$$\\$ Class size of this data = $0.04$$\\ Taking 0.14 as assumed mean (a), d_ i , u_ i , f_ i u_ i are calculated as follows:\\ table \\ From the table, we obtain\\ \Sigma f_i =30\\ \Sigma f_i u_i =-31$$\\$ Mean $\bar{x} =a+(\dfrac{\Sigma f_i u_i}{\Sigma f_i})*h\\ =0.14+(\dfrac{-31}{30})(0.04)\\ =0.14-0.04133\\ =0.099 ppm$$\\ Therefore, mean concentration of SO_ 2 in the air is 0.099 ppm. 8 A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent. ##### Solution : To find the class mark of each interval, the following relation is used.\\ X_i =\dfrac{Upper \ class \ limit \ + \ Lower \ class \ limit}{2}$$\\$ Taking $17$ as assumed mean $(a), d_ i$ and $f _i d_ i$ are calculated as follows:$\\$ table $\\$ From the table, we obtain$\\$ $\Sigma f_i=40\\ \Sigma f_i d_i=181$$\\ Mean \bar{x} =a+(\dfrac{\Sigma f_i d_i}{\Sigma f_i})\\ =17+(\dfrac{-181}{40})\\ 17-4.525\\ 12.48$$\\$ Therefore, the mean number of days is $12.48$ days for which a student was absent.

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

##### Solution :

To find the class marks, the following relation is used.$\\$ Class size (h) for this data =$10$ Taking $70$ as assumed mean $(a), d _i , u_ i$ an d $f_ i u _i$ are calculated as follows.$\\$ table $\\$ From the table, we obtain$\\$ $\Sigma f_i=35\\ \Sigma f_iu_i=-2$$\\ Mean \bar{x} =a+(\dfrac{\Sigma f_i u_i}{\Sigma f_i})*h\\ =70+(\dfrac{-2}{35})*(10)\\ 70-\dfrac{20}{35}\\ =69.43$$\\$ Therefore, mean literacy rate is$69.43 \per .$

10   The following table shows the ages of the patients admitted in a hospital during a year: Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

##### Solution :

To find the class marks (xi), the following relation is used.$\\$ $X_i=\dfrac{Upper class limit+Lower class limit}{2}$$\\ Taking 30 as assumed mean (a), d _i and f_ i d_ i are calculated as follows.\\ From the table, we obtain\\ \Sigma f_i=80\\ \Sigma f_i d_i=430\\ Mean \bar{x}=a+(\dfrac{\Sigma f_i d_i}{\Sigma f_i})\\ =30+(\dfrac{430}{80})\\ =35.38$$\\$ 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$$\\ Frequency (f _1 ) of modal class = 23$$\\$ Class size (h) = $10$$\\ Frequency (f_ 0 ) of class preceding the modal class = 21$$\\$ Frequency $(f _2 )$ of class succeeding the modal class = $14$$\\ Mode =l+(\dfrac{f_1-f_0}{2f_1-f_0-f_2})*h\\ =35+(\dfrac{23-21}{2(23)-21-14})*10\\ 35+\dfrac{20}{11}\\ 35+1.81\\ 36.8$$\\$ Mode is $36.8$. It represents that the age of maximum number of patients admitted in hospital was $36.8$ years.

11   The following data gives the information on the observed lifetimes (in hours) of $225$ electrical components:$\\$ Determine the modal lifetimes of the components.

##### Solution :

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$$\\ Frequency (f _1 ) of modal class = 61$$\\$ Frequency $(f_ 0 )$ of class preceding the modal class = $52$$\\ Frequency (f_ 2 ) of class succeeding the modal class = 38$$\\$ Class size (h) =$20$$\\ Mode =l+(\dfrac{f_1-f_0}{2f_1-f_0-f_2})*h\\ =60+(\dfrac{61-52}{2(61)-52-38})*(20)\\ =60+[\dfrac{9}{122-90}](20)\\ =360+(\dfrac{9*20}{32})\\ =65.625$$\\$ Therefore, modal lifetime of electrical components is $65.625$ hours.

12   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.

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 $(f _1 )$ of modal class = $40$.$\\$ Frequency $(f _0 )$ of class preceding modal class = $24$.$\\$ Frequency $(f _2 )$ of class succeeding modal class = $33$.$\\$ Class size (h) = $500$$\\ Mode =l+(\dfrac{f_1-f_0}{2f_1-f_0-f_2})*h\\ =1500+(\dfrac{40-24}{2(40)-24-33})*500\\ =1500+[\dfrac{16}{80-57}]*500\\ 1500+\dfrac{8000}{23}\\ 1847.826 \simeq 1847.83$$\\$ Therefore, modal monthly expenditure was $Rs 1847.83.$ To find the class mark, the following relation is used.$\\$ Class mark=$\dfrac{Upper class limit+Lower class limit}{2}$$\\ Class size (h) of the given data = 500 Taking 2750 as assumed mean (a), d_ i , u _i an d f _i u _i are calculated as follows.\\ From the table, we obtain\\ \Sigma f_i=200\\ \Sigma f_i u_i=-35\\ \bar{x}(mean) a+(\dfrac{\Sigma f_i u_i}{\Sigma f_i})*h\\ =2750+(\dfrac{-35}{200})*500\\ 2662.5 Therefore, mean monthly expenditure was Rs 2662.50 13 The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two Measures. ##### Solution : It can be observed from the given data that the maximum class frequency is 10 belonging to class interval 30 - 35.$$\\$ Therefore, modal class = $30 - 35$$\\ Class size (h) = 5$$\\$ Lower limit (l) of modal class = $30$$\\ Frequency (f _1 ) of modal class = 10$$\\$ Frequency $(f _0 )$ of class preceding modal class =$$$\\9 Frequency (f_ 2 ) of class succeeding modal class = 3$$\\$ $Mode =l+(\dfrac{f_1-f_0}{2f_1-f_0-f_2})*h\\ 30+(\dfrac{10-9}{2(10)-9-3})*5\\ 30+[\dfrac{1}{20-12}]5\\ =30.6\\ Mode =30.6$$\\ It represents that most of the states/U.T have a teacher-student ratio as 30.6. To find the class marks, the following relation is used.\\ Class mark=\dfrac{Upper class limit+Lower class limit}{2}$$\\$ Taking $32.5$ as assumed mean $(a), d _i , u_ i$ an d $f _i u _i$ are calculated as follows.$\\$ Mean $\bar{x}=a+(\dfrac{\Sigma f_i u_i}{\Sigma f_i})h\\ 32.5+(\dfrac{-23}{35})*5\\ =32.5-3.28=29.22$$\\ Therefore, mean of the data is 29.2.It represents that on an average, teacher-student ratio was 29.2. 14 The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.Find the mode of the data. ##### Solution : From the given data, it can be observed that the maximum class frequency is 18$$\\$ belonging to class interval $4000 - 5000.$$\\ Therefore, modal class = 4000 - 5000$$\\$ Lower limit (l) of modal class = $4000$ Frequency $(f _1 )$ of modal class = $18$ Frequency $(f _0 )$ of class preceding modal class = $4$$\\ Frequency (f _2 ) of class succeeding modal class = 9$$\\$Class size (h) =$1000$$\\ Mode =l+(\dfrac{f_1-f_0}{2f_1-f_0-f_2})h\\ 4000+(\dfrac{18-4}{2(18)-4-9})*1000\\ 4000+608.695\\ =4608.695$$\\$Therefore, mode of the given data is $4608.7$ runs

15   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:

##### Solution :

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$$\\ Frequency (f _1 ) of modal class = 20$$\\$ Frequency $(f _0 )$ of class preceding modal class = $12$$\\ Frequency (f _2 ) of class succeeding modal class = 11$$\\$ Class size = $10$$\\ Mode =l+(\dfrac{f_1-f_0}{2f_1-f_0-f_2})h\\ 40+(\dfrac{20-12}{2(20)-12-11})10\\ 40+\dfrac{80}{17}\\ 44.7$$\\$Therefore, mode of this data is $44.7$ cars.

16   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.

##### Solution :

To find the class marks, the following relation is used.$\\$ Class mark $X_i=\dfrac{Upper class limit+Lower class limit}{2}$$\\ Taking 135 as assumed mean (a), d_ i , u_ i an d f_ i u _i are calculated according to step deviation method as follows :\\From the table, we obtain\\ \Sigma f_i=68\\ \Sigma f_i u_i=7\\ Class \ size =20\\ Mean \bar{x}=a+(\dfrac{\Sigma f_i u_i}{\Sigma f_i})h\\ 135+\dfrac{7}{68}20\\ 137.058$$\\$ 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$$\\ Class size (h) = 20 Frequency (f _1 ) of modal class = 20$$\\$ Frequency $(f_ 0 )$ of class preceding modal class = $13$ Frequency $(f _2 )$ of class succeeding the modal class = $14$$\\ Mode = l +(\dfrac{f_1-f_0}{2f_1-f_0-f_2})h\\ 125+(\dfrac{20-13}{2(20)-13-14})20\\ 125+\dfrac{140}{13}=135.76$$\\$ To find the median of the given data, cumulative frequency is calculated as follows.$\\$ From the table, we obtain $n = 68$$\\ Cumulative frequency (cf) just greater than is 42, belonging to interval 125 - 145$$\\$. Therefore, median class = $125 - 145$$\\ Lower limit (l) of median class = 125$$\\$ Class size (h) = $20$$\\ Frequency (f) of median class = 20$$\\$Cumulative frequency (cf) of class preceding median class =$22$$\\ Median=l+(\dfrac{\dfrac{n}{2}-cf}{f})h\\ =125+(\dfrac{34-22}{20})20\\ =125+12=137$$\\$ Therefore, median, mode, mean of the given data is $137, 135.76,$ and $137.05$ respectively. The three measures are approximately the same in this case.

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

##### Solution :

The cumulative frequency for the given data is calculated as follows. 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$$\\ Class size (h) = 10$$\\$ $Median =l+(\dfrac{\dfrac{n}{2}-cf}{f})h\\ 28.5=20+(\dfrac{\dfrac{60}{2}-(5+x)}{20})10\\ 8.5=(\dfrac{25-x}{2})\\ 17=25-x\\ x=8$$\\ From equation (1), 8 + y = 15\\ y = 7$$\\$ Hence, the values of $x$ and $y$ are $8$ and $7$ respectively

18   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.

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 topersons 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$\dfrac{n}{2}(\dfrac{100}{2}=50) \ 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$$\\ Cumulative frequency (cf) of class preceding median class = 45$$\\$ $Median=l+(\dfrac{\dfrac{n}{2}-cf}{f})h\\ 35+(\dfrac{50-45}{33})5\\ 35+\dfrac{25}{33}\\ 35.76$$\\ Therefore, median age is 35.76 years. 19 The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:Find the median length of the leaves. ##### Solution : The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore, \dfrac{1}{2}= 0.5 has to be added and subtracted to upper class limits and lower class limits respectively.\\ Continuous class intervals with respective cumulative frequencies can be represented as follows.\\ From the table, it can be observed that the cumulative frequency just greater than \dfrac{n}{2} is 29belonging to class interval 144.5 - 153.5.$$\\$ 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$$\\ Median= l+(\dfrac{\dfrac{n}{2}-cf}{f})h\\ 144.5+(\dfrac{20-17}{12})9\\ 146.75$$\\$Therefore, median length of leaves is $146.75 mm.$

20   The following table gives the distribution of the life time of $400$ neon lamps:Find the median life time of a lamp.

##### Solution :

The cumulative frequencies with their respective class intervals are as follows.$\\$ It can be observed that the cumulative frequency just greater than $\dfrac{n}{2} \ is 216$,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(\dfrac{\dfrac{n}{2}-cf}{f})h\\ =3000+(\dfrac{200-130}{86})500\\ =3000+\dfrac{70*500}{86}\\ =3406.976$$\\$ Therefore, median life time of lamps is $3406.98$ hours.

21   $100$ 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.

##### Solution :

The cumulative frequencies with their respective class intervals are as follows.$\\$ It can be observed that the cumulative frequency just greater than$\dfrac{n}{2} \ is 76$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+(\dfrac{\dfrac{n}{2}-cf}{f})h\\ =7+(\dfrac{50-36}{40})3\\ 8.05$$\\ To find the class marks of the given class intervals, the following relation is used.\\ Class mark= \dfrac{Upper class limit+Lower class limit}{2}$$\\$ Taking 11.5 as assumed mean $(a), d_ i , u _i$ an d $f_ i u_ i$ are calculated according to step deviation method as follows.$\\$ From the table, we obtain$\\$ $\Sigma f_i u_i=-106\\ \Sigma f_i=100\\ Mean , \bar{x} =a+(\dfrac{\Sigma f_i u_i}{\Sigma f_i})h\\ 11.5+(\dfrac{-106}{100})3\\ 11.5-3.18=8.32$$\\ The data in the given table can be written as\\ From the table, it can be observed that the maximum class frequency is 40$$\\$ belonging to class interval $7 - 10.$$\\ Modal class = 7 -10$$\\$ Lower limit (l) of modal class =$7 $$\\Class size (h) = 3$$\\$ Frequency $(f _1 )$ of modal class = $40$$\\ Frequency (f _0 ) of class preceding the modal class = 30$$\\$ Frequency $(f _2 )$ of class succeeding the modal class = $16$$\\ Mode =l+(\dfrac{f_1-f_0}{2f_1-f_0-f_2})h\\ =7+[\dfrac{40-30}{2*40-30-16}]3\\ =7+\dfrac{10}{34}3\\ 7.88$$\\$Therefore, median number and mean number of letters in surnames is $8.05$ and $8.32$ respectively while modal size of surnames is $7.88.$

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

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The cumulative frequencies with their respective class intervals are as follows$\\$ Cumulative frequency just greater than$\dfrac{n}{2} \ is 19$,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$$\\ Class size (h) = 5$$\\$ $Median=l+\dfrac{\dfrac{n}{2}-cf}{f}h\\ =55+(\dfrac{15-13}{6})5\\ =55+\dfrac{10}{6}\\ =56.67$$\\ Therefore, median weight is 56.67 kg. 23 The following distribution gives the daily income of 50 workers of a factory.\\ \\ Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive. ##### Solution : The frequency distribution table of less than type is as follows: Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows: 24 During the medical check-up of 35 students of a class, their weights were recorded as follows: Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula. ##### Solution : The given cumulative frequency distributions of less than type areTaking upper class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be drawn as follows.\\ Here,n=35\\ S0 \dfrac{n}{2}=17.5$$\\$Mark the point A whose ordinate is $17.5$ and its x-coordinate is $46.5$.$\\$ Therefore, median of this data is $46.5.$$\\ It can be observed that the difference between two consecutive upper class limits is\\ 2. The class marks with their respective frequencies are obtained as below.\\ The cumulative frequency just greater than\dfrac{n}{2} \ is 28belonging to class interval 46 - 48.$$\\$ 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$$\\ Median=l+(\dfrac{\dfrac{n}{2}-cf}{f})h\\ =46+(\dfrac{17.5-14}{14})2\\ 46+\dfrac{3.5}{7}\\ 46.5$$\\$ Therefore, median of this data is $46.5$.$\\$ Hence, the value of median is verified.

25   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.

##### Solution :

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.