NCERT Solutions Class 10 Maths Chapter 14

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.

$\begin{array}{l}\text{Here, we observe that class marks and frequencies are small}\\ \text{quantities.}\\ \text{So, we use direct method to compute the mean and proceed}\\ \text{as below.}\end{array}$

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}=\frac{162}{20}=8.1\\ \text{Therefore, mean number of plants per house is 8.1.}\end{array}$

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.

$\begin{array}{l}\text{We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=20\\ \text{Here, we take assumed mean(a)}=550\text{and proceed as below.}\end{array}$

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=550+\left(\frac{-12}{50}\right)\times 20=545.2\end{array}$

$\begin{array}{l}\text{Therefore, mean daily wages of the workers is}₹545.20\end{array}$

Q.3 The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.

Ans.

$\begin{array}{l}\text{We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=2\\ \text{Given mean(a)}=18\\ \text{We proceed as below to find}{\mathrm{d}}_{\mathrm{i}},{\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}\text{.}\end{array}$

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ \text{or}18=18+\frac{2\mathrm{f}-40}{44+\mathrm{f}}\\ \text{or \hspace{0.17em}}\mathrm{f}=20\\ \text{Therefore, missing frequency}\mathrm{f}\text{is}20.\end{array}$

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.

$\begin{array}{l}\text{We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=3\\ \text{Here, we take assumed mean(a)}=75.5\text{and proceed as below}\\ \text{to find}{\mathrm{d}}_{\mathrm{i}},{\mathrm{u}}_{\mathrm{i}}\text{and}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=75.5+\left(\frac{4}{30}\right)\times 3=75.9\\ \text{Therefore, mean heartbeats per minute is}75.9.\end{array}$

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.

$\begin{array}{l}\text{The given class intervals are not continuous. So, we add 0.5 to upper class limit and subtract 0.5 to lower class limit}\\ \text{of each interval. We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=3\\ \text{Here, we take assumed mean(a)}=57\text{and proceed as below}\\ \text{to find}{\mathrm{d}}_{\mathrm{i}},{\mathrm{u}}_{\mathrm{i}}\text{and}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=57+\left(\frac{25}{400}\right)\times 3=57.1875\approx 57.19\\ \text{Therefore, mean number of mangoes is}57.19.\\ \text{We have chosen step deviation method as values of}{\mathrm{f}}_{\mathrm{i}}\text{,}{\mathrm{d}}_{\mathrm{i}}\text{are big}\\ \text{and there is a common multiple between all}{\mathrm{d}}_{\mathrm{i}}.\end{array}$

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.

$\begin{array}{l}\text{We find class mark for each interval by using the following relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=50\\ \text{Here, we take assumed mean(a)}=225\text{and proceed as below}\\ \text{to find}{\mathrm{d}}_{\mathrm{i}},{\mathrm{u}}_{\mathrm{i}}\text{and}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=225+\left(\frac{-7}{25}\right)\times 50=211\\ \text{Therefore, mean daily expenditure is}₹211.\end{array}$

Q.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 SO₂ in the air.

Ans.

$\begin{array}{l}\text{We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=0.04\\ \text{Here, we take assumed mean(a)}=0.14\text{and proceed as below}\\ \text{to find}{\mathrm{d}}_{\mathrm{i}},{\mathrm{u}}_{\mathrm{i}}\text{and}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=0.14+\left(\frac{-31}{30}\right)\times 0.04\\ \approx 0.099\text{ppm}\\ {\text{Therefore, mean concentration of SO}}_{\text{2}}\text{in the air is}0.099\text{ppm}.\end{array}$

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.

$\begin{array}{l}\text{We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Assumed mean(a)}=17\\ \text{We proceed as below to find}{\mathrm{d}}_{\mathrm{i}},{\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}\text{.}\end{array}$

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ =17+\frac{-181}{40}\\ =12.475\approx 12.48\\ \text{Therefore, mean number of days for which a student was absent is}12.48.\end{array}$

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

Ans.

$\begin{array}{l}\text{We find class mark for each interval by using the following relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=10\\ \text{Here, we take assumed mean(a)}=70\text{and proceed as below}\\ \text{to find}{\mathrm{d}}_{\mathrm{i}},{\mathrm{u}}_{\mathrm{i}}\text{and}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=70+\left(\frac{-2}{35}\right)\times 10=69.43\\ \text{Therefore, mean literacy rate is}69.43.\end{array}$

Q.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.

Ans.

$\begin{array}{l}\text{We find class mark for each interval by using the following relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Assumed mean(a)}=30\\ \text{We proceed as below to find}{\mathrm{d}}_{\mathrm{i}},{\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}\text{.}\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} Total}=80\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} Total}=430\\ \text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ =30+\frac{430}{80}\\ =35.375\approx 35.38\\ \text{Therefore, mean of the given data is}35.38\text{years}.\\ \text{Modal class is}35-45.\\ \mathrm{l}=35,{\mathrm{f}}_1=23,{\mathrm{f}}_0=21,{\mathrm{f}}_2=14,\mathrm{h}=10\end{array}$

$\begin{array}{l}\text{Mode}=\mathrm{l}+\left(\frac{{\mathrm{f}}_1-{\mathrm{f}}_0}{2{\mathrm{f}}_1-{\mathrm{f}}_0-{\mathrm{f}}_2}\right)\times \mathrm{h}\\ =35+\left(\frac{23-21}{2\times 23-21-14}\right)\times 10\\ =36.8\\ \text{Mode is 36.8 which represents that maximum number}\\ \mathrm{of}\mathrm{patients}\mathrm{admitted}\mathrm{in}\mathrm{hospital}\mathrm{are}\mathrm{of}36.8\mathrm{years}.\end{array}$

$\begin{array}{l}\text{While on average the age of a patient admitted to the}\\ \mathrm{hospital}\mathrm{is}35.38\mathrm{years}\end{array}$

Q.11 The following data gives the information on the observed lifetimes (in hours) of 225 electrical components.

Determine the modal lifetimes of the components.

Ans.

$\begin{array}{l}\text{From the given data, we have}\\ \mathrm{l}=60,{\mathrm{f}}_1=61,{\mathrm{f}}_0=52,{\mathrm{f}}_2=38,\mathrm{h}=20\\ \text{Mode}=\mathrm{l}+\left(\frac{{\mathrm{f}}_1-{\mathrm{f}}_0}{2{\mathrm{f}}_1-{\mathrm{f}}_0-{\mathrm{f}}_2}\right)\times \mathrm{h}\\ =60+\left(\frac{61-52}{2\times 61-52-38}\right)\times 20\\ =65.625\\ \text{So, modal lifetime of electrical components is 65.625 hours.}\end{array}$

Q.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.

Ans.

$\begin{array}{l}\text{We find class mark for each interval by using the following relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Assumed mean(a)}=2750\\ \text{We proceed as below to find}{\mathrm{d}}_{\mathrm{i}},{\mathrm{u}}_{\mathrm{i}},{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\times \mathrm{h}\\ =2750+\frac{-35}{200}\times 500\\ =2662.5\\ \text{Therefore, mean monthly expenditure is}₹2662.5.\\ \text{From the given data, we have}\\ \mathrm{l}=1500,{\mathrm{f}}_1=40,{\mathrm{f}}_0=24,{\mathrm{f}}_2=33,\mathrm{h}=500\\ \text{Mode}=\mathrm{l}+\left(\frac{{\mathrm{f}}_1-{\mathrm{f}}_0}{2{\mathrm{f}}_1-{\mathrm{f}}_0-{\mathrm{f}}_2}\right)\times \mathrm{h}\\ =1500+\left(\frac{40-24}{2\times 40-24-33}\right)\times 500\\ =1847.826\approx 1847.83\\ \text{So, modal monthly expenditure is}₹1847.83.\end{array}$

Q.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.

Ans.

$\begin{array}{l}\text{From the given data, we have}\\ \mathrm{l}=30,{\mathrm{f}}_1=10,{\mathrm{f}}_0=9,{\mathrm{f}}_2=3,\mathrm{h}=5\\ \text{Mode}=\mathrm{l}+\left(\frac{{\mathrm{f}}_1-{\mathrm{f}}_0}{2{\mathrm{f}}_1-{\mathrm{f}}_0-{\mathrm{f}}_2}\right)\times \mathrm{h}\\ =30+\left(\frac{10-9}{2\times 10-9-3}\right)\times 5\\ =30.625\approx 30.6\\ \text{Interpretation:}\\ \text{Most of the states/UT have a teacher student ratio as 30.6.}\\ \text{Now, we find class mark for each interval by using the}\\ \text{following relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Assumed mean(a)}=32.5\\ \text{We proceed as below to find}{\mathrm{d}}_{\mathrm{i}},{\mathrm{u}}_{\mathrm{i}},{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\times \mathrm{h}\\ =32.5+\frac{-23}{35}\times 5\\ =29.2\\ \text{Interpretation:}\\ \text{Average teacher student ratio of the states/UT is}29.2.\end{array}$

Q.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.

Ans.

$\begin{array}{l}\text{From the given data, we have}\\ \mathrm{l}=4000,{\mathrm{f}}_1=18,{\mathrm{f}}_0=4,{\mathrm{f}}_2=9,\mathrm{h}=1000\\ \text{Mode}=\mathrm{l}+\left(\frac{{\mathrm{f}}_1-{\mathrm{f}}_0}{2{\mathrm{f}}_1-{\mathrm{f}}_0-{\mathrm{f}}_2}\right)\times \mathrm{h}\\ =4000+\left(\frac{18-4}{2\times 18-4-9}\right)\times 1000\\ =4608.695\\ \text{Mode of the given data is}4608.7\; (\mathrm{approx}.)\end{array}$

Q.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.

Ans.

$\begin{array}{l}\text{From the given data, we have}\\ \mathrm{l}=40,{\mathrm{f}}_1=20,{\mathrm{f}}_0=12,{\mathrm{f}}_2=11,\mathrm{h}=10\\ \text{Mode}=\mathrm{l}+\left(\frac{{\mathrm{f}}_1-{\mathrm{f}}_0}{2{\mathrm{f}}_1-{\mathrm{f}}_0-{\mathrm{f}}_2}\right)\times \mathrm{h}\\ =40+\left(\frac{20-12}{2\times 20-12-11}\right)\times 10\\ =44.7\\ \therefore \text{Mode of the given data is}44.7.\end{array}$

Q.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.

Ans.

$\begin{array}{l}\text{We find class mark for each interval by using the}\\ \text{following relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Assumed mean(a)}=135\\ \text{We proceed as below to find}{\mathrm{d}}_{\mathrm{i}},{\mathrm{u}}_{\mathrm{i}},{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\times \mathrm{h}\\ =135+\frac{7}{68}\times 20\\ =137.058\\ \text{Modal class}=125-145\\ \text{From the given data, we have}\\ \mathrm{l}=125,{\mathrm{f}}_1=20,{\mathrm{f}}_0=13,{\mathrm{f}}_2=14,\mathrm{h}=20\\ \text{Mode}=\mathrm{l}+\left(\frac{{\mathrm{f}}_1-{\mathrm{f}}_0}{2{\mathrm{f}}_1-{\mathrm{f}}_0-{\mathrm{f}}_2}\right)\times \mathrm{h}\\ =125+\left(\frac{20-13}{2\times 20-13-14}\right)\times 20\\ =135.76\\ \text{We know that}\\ \text{3 Median}=\text{Mode}+\text{2 Mean}\\ \text{or Median}=\frac{135.76+2\times 137.058}{3}=136.625\end{array}$

We observe all these three measures are approximately same.

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

Ans.

We find cumulative frequency of the given data as below.

Here,
n = 60
or 45 + x + y = 60
or x + y = 15 …(1)
Median of data is given as 28.5 that lies in the interval 20 – 30.
So, median class = 20 – 30

$\begin{array}{l}\text{From the given data, we have}\\ \mathrm{l}=20,\\ \text{Cumulative frequency (}\mathrm{cf}\text{) of the class preceding the}\\ \text{median class}=5+\mathrm{x},\\ \text{Frequency of median class (}\mathrm{f})=20,\\ \text{Class size (}\mathrm{h})=10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Median}=\mathrm{l}+\left(\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right)\times \mathrm{h}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}28.5=20+\left(\frac{\frac{60}{2}-5-\mathrm{x}}{20}\right)\times 10\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=8\\ \text{On putting this value of x in equation (1), we get}\\ \mathrm{x}+\mathrm{y}=15\\ \text{or 8}+\mathrm{y}=15\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=15-8=7\\ \text{Hence, values of x and y are 8 and 7 respectively.}\end{array}$

Q.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.

Ans.

$\begin{array}{l}\text{We write the class interval, number of policy holders (}{\mathrm{f}}_{\mathrm{i}}),\text{and cumulative frequency (}\mathrm{cf}\text{) as below in the following table}\\ \text{on the basis of the given information.}\end{array}$

$\begin{array}{l}\text{We have,}\mathrm{n}=100.\\ \therefore \frac{\mathrm{n}}{2}=\frac{100}{2}=50\\ \text{Cumulative frequency (}\mathrm{cf}\text{) just greater than\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{n}}{2}=\text{50 is 78}\\ \text{which belongs to interval 35}-40.\\ \therefore \text{Median class}=35-40\\ \text{Now, we have}\\ \mathrm{l}=35,\\ \text{Cumulative frequency (}\mathrm{cf}\text{) of the class preceding the}\\ \text{median class}=45,\\ \text{Frequency of median class (}\mathrm{f})=33,\end{array}$

$\begin{array}{l}\text{Class size (}\mathrm{h})=5\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Median}=\mathrm{l}+\left(\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right)\times \mathrm{h}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Median\hspace{0.17em}}=35+\left(\frac{\frac{100}{2}-45}{33}\right)\times 5\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=35.76\\ \text{Hence, median age is}35.76\text{years.}\end{array}$

Q.19 The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, 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)

Ans.

$\begin{array}{l}\text{The given class intervals are not continuous. We convert it to continuous class intervals by subtracting 0.5 from each}\\ \text{of lower boundary and adding 0.5 in each of upper boundary.}\\ \\ \text{We write the class interval, number of leaves (}{\mathrm{f}}_{\mathrm{i}}),\text{and cumulative frequency (}\mathrm{cf}\text{) as below in the following table}\\ \text{on the basis of the given information.}\end{array}$

$\begin{array}{l}\text{We have,}\mathrm{n}=40.\\ \therefore \frac{\mathrm{n}}{2}=\frac{40}{2}=20\\ \text{Cumulative frequency (}\mathrm{cf}\text{) just greater than\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{n}}{2}=\text{20 is 29}\\ \text{which belongs to interval}144.5-153.5.\\ \therefore \text{Median class}=144.5-153.5\\ \text{Now, we have}\\ \mathrm{l}=144.5,\\ \text{Cumulative frequency (}\mathrm{cf}\text{) of the class preceding the}\\ \text{median class}=17,\\ \text{Frequency of median class (}\mathrm{f})=12,\\ \text{Class size (}\mathrm{h})=9\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Median}=\mathrm{l}+\left(\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right)\times \mathrm{h}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Median\hspace{0.17em}}=144.5+\left(\frac{\frac{40}{2}-17}{12}\right)\times 9\\ \text{or\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}Median\hspace{0.17em}}=146.75\\ \text{Hence, median length of leaves is}146.75\text{mm.}\end{array}$

Q.20 The following table gives the distribution of the life time of 400 neon lamps:

Find the median life time of a lamp.

Ans.

$\begin{array}{l}\text{We write the cumulative frequency (}\mathrm{cf}\text{) as below in the}\\ \text{following table on the basis of the given information.}\end{array}$

$\begin{array}{l}\text{We have,}\mathrm{n}=400.\\ \therefore \frac{\mathrm{n}}{2}=\frac{400}{2}=200\\ \text{Cumulative frequency (}\mathrm{cf}\text{) just greater than\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{n}}{2}=\text{200 is 216}\\ \text{which belongs to interval}3000-3500.\\ \therefore \text{Median class}=3000-3500\\ \text{Now, we have}\\ \mathrm{l}=3000,\\ \text{Cumulative frequency (}\mathrm{cf}\text{) of the class preceding the}\\ \text{median class}=130,\\ \text{Frequency of median class (}\mathrm{f})=86,\\ \text{Class size (}\mathrm{h})=500\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Median}=\mathrm{l}+\left(\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right)\times \mathrm{h}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Median\hspace{0.17em}}=3000+\left(\frac{\frac{400}{2}-130}{86}\right)\times 500\\ \text{or\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}Median\hspace{0.17em}}=3406.98\\ \text{Hence, median life time of a lamp is}3406.98\text{hours.}\end{array}$

Q.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.

Ans.

$\begin{array}{l}\text{We find the cumulative frequency (}\mathrm{cf}\text{) as below in the}\\ \text{following table on the basis of the given information.}\end{array}$

$\begin{array}{l}\text{Total(}\mathrm{n}\text{)}=100\\ \text{We have,}\mathrm{n}=100.\\ \therefore \frac{\mathrm{n}}{2}=\frac{100}{2}=50\\ \text{Cumulative frequency (}\mathrm{cf}\text{) just greater than\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{n}}{2}=\text{50 is 76}\\ \text{which belongs to interval}7-10.\\ \therefore \text{Median class}=7-10\end{array}$

$\begin{array}{l}\text{Now, we have}\\ \mathrm{l}=7,\\ \text{Cumulative frequency (}\mathrm{cf}\text{) of the class preceding the}\\ \text{median class}=36,\\ \text{Frequency of median class (}\mathrm{f})=40,\\ \text{Class size (}\mathrm{h})=3\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Median}=\mathrm{l}+\left(\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right)\times \mathrm{h}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Median\hspace{0.17em}}=7+\left(\frac{\frac{100}{2}-36}{40}\right)\times 3\\ \text{or\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}Median\hspace{0.17em}}=8.05\\ \therefore \text{Median number of letters in the surnames is 8.05.}\\ \text{Now, we find class mark for each interval by using the}\\ \text{following relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Assumed mean(a)}=11.5\end{array}$

We proceed as below to find d_i, u_i, f_iu_i .

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\times \mathrm{h}\\ =11.5+\frac{-106}{100}\times 3\\ =8.32\\ \text{We know that}\\ \text{3 Median}=\text{Mode}+\text{2 Mean}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3}\times 8.05=\text{Mode}+\text{2}\times 8.32\\ \text{or Mode}=7.51\\ \text{Hence, mean number of letters in the surnames is}8.32\\ \text{and modal size of the surnames is}7.51.\end{array}$

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

Ans.

$\begin{array}{l}\text{We find the cumulative frequency (}\mathrm{cf}\text{) as below in the}\\ \text{following table on the basis of the given information.}\end{array}$

$\begin{array}{l}\text{We have,}\mathrm{n}=30.\\ \therefore \frac{\mathrm{n}}{2}=\frac{30}{2}=15\\ \text{Cumulative frequency (}\mathrm{cf}\text{) just greater than\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{n}}{2}=15\text{is 19}\\ \text{which belongs to interval 55}-60.\\ \therefore \text{Median class}=55-60\\ \text{Now, we have}\\ \mathrm{l}=55,\\ \text{Cumulative frequency (}\mathrm{cf}\text{) of the class preceding the}\\ \text{median class}=13,\\ \text{Frequency of median class (}\mathrm{f})=6,\\ \text{Class size (}\mathrm{h})=5\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Median}=\mathrm{l}+\left(\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right)\times \mathrm{h}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Median\hspace{0.17em}}=55+\left(\frac{\frac{30}{2}-13}{6}\right)\times 5\\ \text{or\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}Median\hspace{0.17em}}=56.67\\ \text{Hence, median weight of students is}56.67\text{kg.}\end{array}$

Q.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 graph.

Ans.

$\text{We find less than type cumulative frequency table as below.}$

$\begin{array}{l}\text{Now, taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis we draw its}\\ \text{ogive as following.}\end{array}$

Q.24 During the medical check-up of 35 students of a class, their weights were recorded as follows:

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

Ans.

$\text{Following is the given cumulative frequency distributions of less than type}$

Now locate 17.5 (half of 35) on y-axis and draw a line parallel to x-axis to cut the graph at a point.
Draw a perpendicular from this point to x-axis. The point where this perpendicular meets the x-axis
determines the median of the data. From the graph we see the median is 46.5.

$\begin{array}{l}\text{Now, we mark the point A(17.5, 46.5). So median of the given data is 46.5.}\\ \text{We find class intervals with their respective frequencies as the following.}\end{array}$

$\begin{array}{l}\frac{\mathrm{n}}{2}=\frac{35}{2}=17.5\\ \text{Cumulative frequency just greater that 17.5 is 28 which belongs}\\ \text{to class interval}46-48.\\ \therefore \text{Median class}=46-48\\ \text{Now, we have}\\ \mathrm{l}=46,\\ \text{Cumulative frequency (}\mathrm{cf}\text{) of the class preceding the}\\ \text{median class}=14,\\ \text{Frequency of median class (}\mathrm{f})=14,\\ \text{Class size (}\mathrm{h})=2\end{array}$

$\begin{array}{l}\text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Median}=\mathrm{l}+\left(\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right)\times \mathrm{h}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Median\hspace{0.17em}}=46+\left(\frac{\frac{35}{2}-14}{14}\right)\times 2\\ \text{or\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}Median\hspace{0.17em}}=46.5\\ \text{So median of the given data is}46.5.\\ \text{Hence, value of the median is verified.}\end{array}$

Q.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 its graph.

Ans.

$\text{Following is the cumulative frequency distributions of more than type of the given data..}$

Now, draw the graph by plotting the points : (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16).