NCERT Solutions for Maths Class 10 Chapter 14

Exercise 14.1

Question 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?

Answer:

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

Class mark (xi) \(=\frac{Upper\space class\space limit + Lowwer\space class\space limit}{2}\)

\(x_i\) and \(f_ix_i\) can be calcilated as follows:

From the table, it can be observed that

\(\sum f_i=20\)
\(\sum f_ix_i=162\)
Mean, \(\bar x=\frac{\sum f_ix_i}{\sum f_i}\)
\(\frac{162}{2}=8.1\)
Therefore, mean number of plants per house is \(8.1\).
Here, direct method has been used as the values of class marks (\(x_i\)) and \(f_i\) are small.

Question 2: Consider the following distribution of daily wages of 50 worker of a factory.

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

Answer: To find the class mark for each interval, the following relation is used.

\(x_i\frac{Upper\space class\space limit+Lower\space class\space limit}{2}\)

Class size (h) of this data \(=20\)

Taking 150 as assured mean (a) \(d_i\space m_i\space u_i\) and \(f_i\space c_i\) can be calculated as follows.

From the table, it can be observed that

\(\sum\space f_i=50\)
\(\sum\space f_i\space u_i=-12\)
Mean \(\bar x\space a+(\frac{\sum\space f_i\space u_i}{\sum \space f_i})h\)
\(=150+(\frac{-12}{50})20\)
\(=150-\frac{24}{5}\)
\(=150-4.8\)
\(=145.2\)
Therefore, the mean daily wage of the workers of the factory is Rs. \(145.20\).

Question 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 \(f\).

Answer: 

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

\(x_i=\frac{Upper\space class\space limit+Lower\space class\space limt}{2}\)

Given that, mean pocket allowance, \(\bar x=Rs.18\)

Taking 18 as assured mean (a), \(d_i\) and \(f_i\space d_i\) calculated as follows.

From the table, we obtain

\(\sum f_i=44+f\)

\(\sum f_i\space u_i=2f-40\)

\(\bar x=a+\frac{\sum f_i\space d_i}{\sum f_i}\)

\(18=18+(\frac{2f-40}{44+f})\)

\(0=(\frac{2f-40}{44+f})\)

\(2f-40=0\)

\(2f=40\)

\(f=20\)

Hence, the missing frequency \(f\) is \(20\).

Question 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. Fine the mean heart beats per minute for these women, choosing a suitable method.

Answer:

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

\(x_i=\frac{Upper\space class\space limit+Lower\space class\space limit}{2}\)

Class size, h, of this data \(=3\)

Taking 75.5 as assumed mean (a) \(d_i,\space u_i,\space f_i,\space u_i\) are calculated as follows.

From the table, we obtain

\(\sum f_i=30\)

\(\sum f_i\space u_i=4\)

Mean \(\bar x=a+(\frac{\sum f_i\space u_i}{\sum f_i})\times h\)

\(=75.5+(\frac{4}{30})\times 3\)

\(=75.5+0.4=75.9\)

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

Question 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?

Answer:

It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore,  \(\frac{1}{2}\) has to be added to the upper class limit and  \(\frac{1}{2}\) has to be subtracted from the lower class limit of each interval. Class mark (xi) can be obtained by using the following relation.

\(x_i=\frac{Upper\space class\space limit+Lower\space class\space limit}{2}\)

Class size (h) of this data \(= 3\)

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

It can be observed that

\(\sum f_i=400\)

\(\sum f_i\space u_i=25\)

Mean \(\bar x=a+(\frac{\sum f_i\space u_i}{\sum f_i})\times h\)

\(=57+(\frac{25}{400}\times 3\)

\(=57+\frac{3}{16}=57+0.1875\)

\(57.1875\)

\(=57.19\)

Mean number of mangoes kept in a packing box is \(57.19\).

Step deviation method is used here as the values of \(f_i,\space d_i\) are big and a;sp, there a common multiple between all \(d_i\).

Question 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.

Answer: 

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

\(x_i=\frac{Upper\space class\space limit+Lower\space class \space limit}{2}\)

Class size \(=50\)

Taking \(225\) as assumed mean (a), \(d_i,\space u_i,\space f_i,\space u_i\) are calculated follows.

From the table, we obtain

\(\sum f_i=25\)

\(\sum f_i\space u_-7\)

Mean \(\bar x=a+(\frac{\sum f_i\space u_i}{\sum f_i\space u_i})\times h\)

\(=225+(\frac{-7}{25}\times (5)\)

\(=225-14\)

\(=211\)

Therefore, mean daily expenditure on food is Rs. \(211\).

Question 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:

Find the mean concentration of \(SO_2\) in the air.

Answer: To find the class marks for each interval, the following relation is used.

\(x_i=\frac{Uppe\space class\space limit+Lower\space class\space limi}{2}\)

Class size of this data \(=0.04\)

Taking \(0.14\) as assumed mean (a) \(d_i,\space u_i,\space f_i,\space u_i\) are calculated as follows:

From the table, we obtain

\(\sum f_i=30\)

\(\sum f_i\space u_i=-31\)

Mean \(\bar x=a+(\frac{\sum f_i\space u_i}{\sum f_i})\times h\)

\(0.14+(\frac{-31}{30})(0.04)\)

\(=0.14-0.04133\)

\(=0.0984133\)

\(=0.09867\)

\(=0.099\space ppm\)

Therefore, mean concentration of \(SO_2\) in the air is \(0.099\space ppm\).

Question 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.

Answer: To find the class mark of each interval, the following relation is used.

\(x_i=\frac{Upper\space classs\space limit+Lower\space class\space limit}{2}\)

Takin 17 as assumed mean (a), \(d_i\) and \(f_i\space d_i\) are calculated as follows.

From the table, we obtain

\(\sum f_i=40\)

\(\sum f_i\space d_i=-181\)

Mean, \(\bar x=a+(\frac{\sum f_i\space d_i}{\sum f_i})\)

\(=17+(\frac{-181}{40})\)

\(=17-4.525\)

\(=12.475\)

\(=112.48\)

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

Question 9: The following table gives the litercy rate (in percentage) of 35 cities.

Find the mean litercy rate.

Answer: 

To find the class marks, the following relation is used.

\(x_i=\frac{Upper\space class\space limt+Lower\space class \space limit}{2}\)

Class size (h) for this data \(=10\)

Takin 70 as assumed mean (a), \(d_i\space u_i\) and \(f_i \space u_i\) are calculated follows:

From the table, we obtain

\(\sum f_i=35\)

\(\sum f_i\space u_i=-2\)

Mean \(\bar x=a+(\frac{\sum f_i\space u_i}{\sum f_i})\times h\)

\(=70+(\frac{-2}{35})\times (10)\)

\(=70-\frac{20}{35}\)

\(=70-\frac{20}{35}\)

\(=70-\frac{20}{35}\)

\(=70-\frac{4}{7}\)

\(=70-0.57\)

\(=69.43\)

Therefore, mean literacy rate is \(69.43%\).

Exercise 14.2

Question 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 data given above. Compare and interpret the two measures of central tendency.

Answer:

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

\(x_i=\frac{Upper\space class+Lower \space class \space limit}{2}\)

Taking 30 as assumed mean (a), \(d_i\) and \(f_i\space d_i\) are calculated as follows.

From the table, we obtain

\(\sum f_i=80\)

\(\sum f_i\space d_i=430\)

Mean, \(\bar x=a+\frac{\sum f_i\space d_i}{\sum f_i}\)

\(=30(\frac{430}{80})\)

\(=30+5.375\)

\(=35.38\)

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 \((f_i)\) of modal class \(=23\)

Class \((h)=10\)

Frequency \((f_0)\) of class preceeding the modal class \(=21\)

Frequency \((f_2)\) of class suceeding the modal class \(=14\)

Mode \(=l(\frac{f_i-f_0}{2f_1-f_0-f_2})=\times h\)

\(=35+(\frac{23-21}{2(23)-21-14})\times10\)

\(=35+[\frac{2}{46-35}]\times10\)9365492935

\(=35+\frac{20}{11}\)

\(=35+1.81\)

\(=26.8\)

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

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

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 \((f_1)\) 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 \(=l(\frac{f_i-f_0}{2f_1-f_0-f_2})\times h\)

\(=60+(\frac{61-52}{2(61)-52-38})(20)\)

\(=60+(\frac{9}{122-90})(2)\)

\(=60+(\frac{9\times20}{32})\)

\(=60\frac{90}{16}=60+5.625\)

\(=65.625\)

Therefore, modal lifetime of electrical components is \(65.625\) hours.

Question 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.

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 \((f_1)\) 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 \(l+(\frac{f_1-f_0}{2f_1-f_0-f_2})\times h\)

\(=1500+(\frac{40-24}{2(4)-24-33})\times 500\)

\(=1500+(\frac{16}{80-57})\times 500\)

\(=1500+\frac{8000}{23}\)

\(=1500+347.826\)

\(=1847.826=1847.83\)

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

Class Mark \(=\frac{Upper\space class\space limit+Lower\space class\space limit}{2}\)

Class size \((h)\) of the given data \(=500\)

Taking \(2750\) as assumed mean (a), \(d_i\space u_i\) are calculated as follows.

From the table, we obtain

\(\sum f_i=200\)

\(\sum f_i\space u_i=-35\)

\(\bar x (mean0=a+(\frac{\sum f_i\space u_i}{\sum f_i})\times h\)

\(\bar x=2750+(\frac{-35}{200})\times 500\)

\(=2750-87.5\)

\(=2662.5\)

Therefore, mean monthly expenditure was Rs. \(2662.50.\)

Question 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.

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 \((f_1)\) of modal class \(= 10\)

Frequency \((f_0)\) of class preceding modal class \(= 9\)
Frequency \((f_2)\) of class succeeding modal class \(= 3\)

Mode \(=l(f_i-\frac{f_0}{2f_i-f_0-f_2})\times h\)

\(=30+(\frac{10-9}{2(10)-9-3})\times (5)\)

\(=30+(\frac{1}{20-12})5\)

\(=30+\frac{5}{8}=30.625\)

Mode \(9=30.6\)

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\(=\frac{Upper\space class\space limit+Lower\space class\space limit}{2}\)

Taking \(32.5\) as assumed mean (a), \(d_i,\space u_i,\) and \(f_i\space u_i\) are calculated as follows.

Mean, \(\bar x=a+(\frac{\sum f_i\space u_i}{\sum f_i})h\)

\(=32.5+(\frac{-23}{35})\times 5\)

\(=32.5-\frac{23}{7}=32.5-3.28\)

\(=29.22\)

Therfore, mean of the data is \(29.2\)

It represents that on an average, teacher-student ratio was \(29.2\)

Question 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

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 \((f_1)\) 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 \(=l+(\frac{f_i-f_0}{2f_i-f_0-f_2})\times h\)

\(=4000+(\frac{18-4}{2(18)-4-0})\times 1000\)

\(=4000+(\frac{14000}{23})\)

\(=4000+608.695\)

\(=4608.695\)

Therefore, mode of the given data is \(4608.7\) runs

Question 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:

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 \((f_1)\) 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 \(=l+(\frac{f_1-f_0}{2f_1-f_0-f_2})\times h\)

\(=40+[\frac{20-12}{2(20)-12-11}]\times 10\)

\(=40+(\frac{80}{40-23})\)

\(=40+\frac{80}{17}\)

\(=40+4.7\)

\(=44.7\)

Therefore, mode of this data \(44.7\) cars.

Exercise 14.3

Question 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.

Answer:

To find the class marks, the following relation is used.

Class mark \(=\frac{Upper\space class\space limit+Lower\space limt}{2}\)

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

From the table, we obtain

\(\sum f_i\space u_i=7\)

\(\sum f_i=68\)

Class size \((h)=20\)

Mean, \(\bar x=a+(\frac{\sum f_i\space u_i}{\sum f_i})\times h\)

\(=135+\frac{7}{68}\times 20\)

\(=135+\frac{140}{168}\)

\(=137.058\)

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 \(=l+(\frac{f_i-f_0}{2f_i-f_0-f_2})\times h\)

\(=125+[\frac{20-13}{2(20)-13-14}]\times 20\)

\(=125+\frac{7}{13}\times 20\)

\(=125+\frac{140}{13}=135.76\)

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 (i.e., \(\frac{68}{2}=34)\) is \(42\).

Belonging to interval \(125-145\).

Therfore, 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 \(=1+(\frac{\frac{n}{2}-cf}{f})\times h\)

\(=125+(\frac{34-20}{20})\times 20\)

\(=125+12\)

\(=137\)

Therefore, median, mode, mean of the given data is \(137,\space 135.76\) and \(137.05\) respectively.

The three measures are approximately the same in this case.

Question 2:

If the median of the distribution is given below is \(28.5\) find the values of \(x\) and \(y\).

Answer:

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 \(=1+(\frac{\frac{n}{2}-cf}{f})\times h\)

\(28.20+[\frac{\frac{60}{2}=(5+x)}{20}]\times 10\)

\(⇒8.5(\frac{25-x}{2})\)

\(⇒17=25-x\)

\(⇒x=8\)

From equation (1),

\(⇒8+y=15\)

\(⇒y=7\)

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

Question 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.

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.

Total \((n)\)

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

Cumuulative frequency \((cf)\) just greater than \(\frac{n}{2}(i.e., \frac{100}{2}=50)\) 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 \(=l(\frac{\frac{n}{2}-cf}{f})\times h\)

\(=35+(\frac{50-45}{33})\times 5\)

\(=35+\frac{25}{33}\)

\(=35.76\)

Therefore, median age is \(35.76\) years.

Question 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 (\(117.5 − 126.5, 126.5 − 135.5… 171.5 − 180.5\))

Answer: 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

\(\frac{n}{2}(i.e., \frac{40}{2}=20)\) 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 \((h)=9\)

Frequecncy \((f)\) of median class \(=12\)

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

Mean \(l+(\frac{\frac{n}{2}-cf}{f})\times h\)

\(=144.5+(\frac{20-17}{12})\times h\)

\(=144.5+\frac{9}{4}=146.75\)

Therefore, median length of leaves is \(146.75\) mm.

Question 5:

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

Find the median life time of a lamp.

Answer:

Thecumulative frequencies with their respective class intervals are as follows.

It can be observed that the cumulative frequency just greater than

\(\frac{n}{2}(i.e.,\frac{400}{2}=200)\) 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 \(=l+(\frac{\frac{n}{2}-cf}{f})\times h\)

\(=300+\frac{70\times500}{86}\)

\(=3406.976\)

Thefore, median life time of lamps is \(3406.98\) hours.

Question 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.

Answer: 

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

It can be observed that the cumulative frequency just grreater than \(\frac{n}{2}(i.e., \frac{100}{2}=50)\) is \(76\), belong 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 \(l+(\frac{\frac{n}{2}-cf}{f})\times h\)

\(=7+(\frac{50+46}{40})\times 3\)

\(=7+\frac{14\times 3}{40}\)

\(=8.05\)

To find the class marks of the given class intervals, the following relation is used.

Class mark \(=\frac{Upper\space class\space limit+Lower\space class\space limit}{2}\)

Taking 11.5 as assumed mean (a), \(u_i,\space u_i\) and \(f_i\space u_i\) are calculated according to step deviation method as follows.

From the table, we obtain

\(\sum f_i\space u_i=-106\)

\(\sum f_i=100\)

Mean, \(\bar x=a+(\frac{\sum f_i\space u_i}{\sum f_i})h\)

\(=11.5+(\frac{-106}{100})\times 3\)

\(=11.5-3.18=8.32\)

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 limt \((l)\) of modal class \(=7\)

Class size \((h)=3\)

Frequency \((f_i)\) 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 \(=l+(\frac{f_1-f_0}{2f_1-f_0-f_2})\times h\)

\(=7+[\frac{40-30}{2(40)-30-16}]\times 3\)

\(=7+\frac{10}{34}\times 3\)

\(7+\frac{30}{34}=7.88\)

Therefore, median number and mean number of letters in surnames is \(8.05\) and \(8.32 \) respectively while modal size of surnames is \(7.88.\)

Question 7: The distribution below gives the weights of \(30\) students of a class. Find the median weight of the students.

Answer:

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

Comulative frequency just greater than \(\frac{n}{2}(i.e.,\frac{20}{2}=15)\) is \(19\), belong to class intercal \(55-60\).

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 \(=l+(\frac{\frac{n}{2}-cf}{f})\times h\)

\(=55+(\frac{15-13}{6})\times 5\)

\(=55+\frac{10}{6}\)

\(=56.67\)

Therefore, median weight is \(56.67\) kg.

Exercise 14.4

Question 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.

Answer:

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.

Question 2:

During the medical check-up of 35 students of a class, their weights were recordedas 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.

Answer:

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 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 \(\frac{n}{2}(i.e.,\frac{35}{2}=17.5)\) is \(28\), belonging to class interval \(46-48\).

Median class \(=46-48\).

Lower class limit \((l)\) of median class \(=46\)

Frequenct \((f))\) of median class \(=14\)

Cumulative frequency \((cf)\) of class preceding median class \(=14\)

Class size \((h)=2\)

Median \(=l+(\frac{\frac{n}{2}-cf}{f})\times h\)

\(=46+(\frac{17.5-14}{14})\times 2\)

\(=46+\frac{3.5}{7}\)

\(=46.5\)

Therfore, median of this data is \(46.5\)

Hence, the value of median is verified.

Question 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.

Answer:

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.