1Exercise 13.19 questions
Q.1A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
Number of plants: 0 - 2, 2 - 4, 4 - 6, 6 - 8, 8 - 10, 10 - 12, 12 - 14
Number of houses: 1, 2, 1, 5, 6, 2, 3
Which method did you use for finding the mean, and why?v
SolutionClass marks are $1,3,5,7,9,11,13$. Then $\sum f_i x_i=1(1)+2(3)+1(5)+5(7)+6(9)+2(11)+3(13)=162$ and $\sum f_i=20$. Mean $\bar{x}=\dfrac{\sum f_i x_i}{\sum f_i}=\dfrac{162}{20}=8.1$.
Answer:The mean number of plants per house is $8.1$. The direct method is convenient because the class marks and frequencies are small.
Q.2Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in ₹): 500 - 520, 520 - 540, 540 - 560, 560 - 580, 580 - 600
Number of workers: 12, 14, 8, 6, 10
Find the mean daily wages of the workers of the factory by using an appropriate method.v
SolutionClass marks are $510,530,550,570,590$. Then $\sum f_i x_i=12(510)+14(530)+8(550)+6(570)+10(590)=27260$ and $\sum f_i=50$. Mean $=\dfrac{27260}{50}=545.2$.
Answer:The mean daily wage is $₹545.20$.
Q.3The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency $f$.
Daily pocket allowance (in ₹): 11 - 13, 13 - 15, 15 - 17, 17 - 19, 19 - 21, 21 - 23, 23 - 25
Number of children: 7, 6, 9, 13, $f$, 5, 4v
SolutionClass marks are $12,14,16,18,20,22,24$. The known total frequency is $44$ and the known sum $\sum f_i x_i=752$. Since the mean is 18, $\dfrac{752+20f}{44+f}=18$. Thus $752+20f=792+18f$, so $2f=40$ and $f=20$.
Answer:The missing frequency is $f=20$.
Q.4Thirty 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.
Number of heartbeats per minute: 65 - 68, 68 - 71, 71 - 74, 74 - 77, 77 - 80, 80 - 83, 83 - 86
Number of women: 2, 4, 3, 8, 7, 4, 2v
SolutionClass marks are $66.5,69.5,72.5,75.5,78.5,81.5,84.5$. Then $\sum f_i x_i=2277$ and $\sum f_i=30$. Mean $=\dfrac{2277}{30}=75.9$.
Answer:The mean heartbeats per minute is $75.9$.
Q.5In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
Number of mangoes: 50 - 52, 53 - 55, 56 - 58, 59 - 61, 62 - 64
Number of boxes: 15, 110, 135, 115, 25
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?v
SolutionClass marks are $51,54,57,60,63$. Then $\sum f_i=400$ and $\sum f_i x_i=22875$. Therefore mean $=\dfrac{22875}{400}=57.1875\approx57.19$.
Answer:The mean number of mangoes per box is approximately $57.19$. The assumed mean or step-deviation method is convenient because the frequencies are large.
Q.6The table below shows the daily expenditure on food of 25 households in a locality.
Daily expenditure (in ₹): 100 - 150, 150 - 200, 200 - 250, 250 - 300, 300 - 350
Number of households: 4, 5, 12, 2, 2
Find the mean daily expenditure on food by a suitable method.v
SolutionClass marks are $125,175,225,275,325$. Then $\sum f_i x_i=4(125)+5(175)+12(225)+2(275)+2(325)=5275$ and $\sum f_i=25$. Mean $=\dfrac{5275}{25}=211$.
Answer:The mean daily expenditure on food is $₹211$.
Q.7To find out the concentration of $SO_2$ 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:
Concentration of $SO_2$ (in ppm): 0.00 - 0.04, 0.04 - 0.08, 0.08 - 0.12, 0.12 - 0.16, 0.16 - 0.20, 0.20 - 0.24
Frequency: 4, 9, 9, 2, 4, 2
Find the mean concentration of $SO_2$ in the air.v
SolutionClass marks are $0.02,0.06,0.10,0.14,0.18,0.22$. Then $\sum f_i x_i=2.96$ and $\sum f_i=30$. Mean $=\dfrac{2.96}{30}=0.098666\ldots\approx0.099$ ppm.
Answer:The mean concentration of $SO_2$ is approximately $0.099$ ppm.
Q.8A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Number of days: 0 - 6, 6 - 10, 10 - 14, 14 - 20, 20 - 28, 28 - 38, 38 - 40
Number of students: 11, 10, 7, 4, 4, 3, 1v
SolutionClass marks are $3,8,12,17,24,33,39$. Then $\sum f_i x_i=499$ and $\sum f_i=40$. Mean $=\dfrac{499}{40}=12.475$ days.
Answer:The mean number of days a student was absent is $12.475$ days.
Q.9The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
Literacy rate (in %): 45 - 55, 55 - 65, 65 - 75, 75 - 85, 85 - 95
Number of cities: 3, 10, 11, 8, 3v
SolutionClass marks are $50,60,70,80,90$. Then $\sum f_i x_i=2430$ and $\sum f_i=35$. Mean $=\dfrac{2430}{35}=69.4286\ldots\approx69.43\%$.
Answer:The mean literacy rate is approximately $69.43\%$.
2Exercise 13.26 questions
Q.1The following table shows the ages of the patients admitted in a hospital during a year:
Age (in years): 5 - 15, 15 - 25, 25 - 35, 35 - 45, 45 - 55, 55 - 65
Number of patients: 6, 11, 21, 23, 14, 5
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.v
SolutionFor the mean, class marks are $10,20,30,40,50,60$. Thus $\sum f_i x_i=2830$ and $\sum f_i=80$, so mean $=35.375$. The modal class is $35-45$ with $l=35$, $h=10$, $f_1=23$, $f_0=21$, $f_2=14$. Mode $=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}h=35+\dfrac{2}{11}\times10=36.82$ years approximately.
Answer:Mode $\approx36.82$ years and mean $=35.375$ years. The two values are close; the most common age is about $36.82$ years, while the average age is about $35.38$ years.
Q.2The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :
Lifetimes (in hours): 0 - 20, 20 - 40, 40 - 60, 60 - 80, 80 - 100, 100 - 120
Frequency: 10, 35, 52, 61, 38, 29
Determine the modal lifetimes of the components.v
SolutionThe modal class is $60-80$ with $l=60$, $h=20$, $f_1=61$, $f_0=52$, $f_2=38$. Mode $=60+\dfrac{61-52}{2(61)-52-38}\times20=60+\dfrac{9}{32}\times20=65.625$ hours.
Answer:The modal lifetime is $65.625$ hours.
Q.3The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :
Expenditure (in ₹): 1000 - 1500, 1500 - 2000, 2000 - 2500, 2500 - 3000, 3000 - 3500, 3500 - 4000, 4000 - 4500, 4500 - 5000
Number of families: 24, 40, 33, 28, 30, 22, 16, 7v
SolutionThe modal class is $1500-2000$, so $l=1500$, $h=500$, $f_1=40$, $f_0=24$, $f_2=33$. Mode $=1500+\dfrac{40-24}{2(40)-24-33}\times500=1847.83$ approximately. For the mean, class marks are $1250,1750,2250,2750,3250,3750,4250,4750$; $\sum f_i x_i=532500$ and $\sum f_i=200$, so mean $=2662.5$.
Answer:The modal monthly expenditure is approximately $₹1847.83$, and the mean monthly expenditure is $₹2662.50$.
Q.4The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
Number of students per teacher: 15 - 20, 20 - 25, 25 - 30, 30 - 35, 35 - 40, 40 - 45, 45 - 50, 50 - 55
Number of states / U.T.: 3, 8, 9, 10, 3, 0, 0, 2v
SolutionThe modal class is $30-35$, with $l=30$, $h=5$, $f_1=10$, $f_0=9$, $f_2=3$. Mode $=30+\dfrac{10-9}{20-9-3}\times5=30.625$. For the mean, using class marks $17.5,22.5,27.5,32.5,37.5,42.5,47.5,52.5$, $\sum f_i x_i=1022.5$ and $\sum f_i=35$, so mean $=29.214\ldots$.
Answer:Mode $=30.625$ students per teacher and mean $\approx29.21$ students per teacher. The most common ratio is about $31$ students per teacher, while the average is about $29$ students per teacher.
Q.5The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
Runs scored: 3000 - 4000, 4000 - 5000, 5000 - 6000, 6000 - 7000, 7000 - 8000, 8000 - 9000, 9000 - 10000, 10000 - 11000
Number of batsmen: 4, 18, 9, 7, 6, 3, 1, 1
Find the mode of the data.v
SolutionThe modal class is $4000-5000$, so $l=4000$, $h=1000$, $f_1=18$, $f_0=4$, $f_2=9$. Mode $=4000+\dfrac{18-4}{2(18)-4-9}\times1000=4000+\dfrac{14}{23}\times1000=4608.70$ approximately.
Answer:The mode is approximately $4608.70$ runs.
Q.6A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :
Number of cars: 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50, 50 - 60, 60 - 70, 70 - 80
Frequency: 7, 14, 13, 12, 20, 11, 15, 8v
SolutionThe modal class is $40-50$, so $l=40$, $h=10$, $f_1=20$, $f_0=12$, $f_2=11$. Mode $=40+\dfrac{20-12}{2(20)-12-11}\times10=40+\dfrac{8}{17}\times10=44.71$ approximately.
Answer:The mode is approximately $44.71$ cars.
3Exercise 13.37 questions
Q.1The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption (in units): 65 - 85, 85 - 105, 105 - 125, 125 - 145, 145 - 165, 165 - 185, 185 - 205
Number of consumers: 4, 5, 13, 20, 14, 8, 4v
SolutionFor the mean, class marks give $\sum f_i x_i=9320$ and $\sum f_i=68$, so mean $=137.06$. Cumulative frequencies are $4,9,22,42,56,64,68$; $n/2=34$, so the median class is $125-145$. Median $=125+\dfrac{34-22}{20}\times20=137$. The modal class is also $125-145$: mode $=125+\dfrac{20-13}{2(20)-13-14}\times20=135.77$ approximately.
Answer:Median $=137$ units, mean $\approx137.06$ units, and mode $\approx135.77$ units. The three measures are very close, so the distribution is fairly balanced around about $136$-$137$ units.
Q.2If the median of the distribution given below is 28.5, find the values of $x$ and $y$.
Class interval: 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50, 50 - 60
Frequency: 5, $x$, 20, 15, $y$, 5
Total: 60v
SolutionFrom the total frequency, $5+x+20+15+y+5=60$, so $x+y=15$. Since the median is 28.5, the median class is $20-30$. Here $l=20$, $h=10$, $f=20$, $cf=5+x$ and $n/2=30$. Thus $28.5=20+\dfrac{30-(5+x)}{20}\times10$. This gives $8.5=\dfrac{25-x}{2}$, so $x=8$. Therefore $y=15-8=7$.
Q.3A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
Age (in years): Below 20, Below 25, Below 30, Below 35, Below 40, Below 45, Below 50, Below 55, Below 60
Number of policy holders: 2, 6, 24, 45, 78, 89, 92, 98, 100v
SolutionConvert the less-than cumulative distribution to class frequencies: $18-20:2$, $20-25:4$, $25-30:18$, $30-35:21$, $35-40:33$, $40-45:11$, $45-50:3$, $50-55:6$, $55-60:2$. Here $n=100$, so $n/2=50$. The median class is $35-40$, with $l=35$, $h=5$, $cf=45$, $f=33$. Median $=35+\dfrac{50-45}{33}\times5=35.76$ years approximately.
Answer:The median age is approximately $35.76$ years.
Q.4The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :
Length (in mm): 118 - 126, 127 - 135, 136 - 144, 145 - 153, 154 - 162, 163 - 171, 172 - 180
Number of leaves: 3, 5, 9, 12, 5, 4, 2
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.)v
SolutionAfter converting to continuous classes, cumulative frequencies are $3,8,17,29,34,38,40$. Here $n=40$, so $n/2=20$. The median class is $144.5-153.5$, with $l=144.5$, $h=9$, $cf=17$, $f=12$. Median $=144.5+\dfrac{20-17}{12}\times9=146.75$ mm.
Answer:The median length is $146.75$ mm.
Q.5The following table gives the distribution of the life time of 400 neon lamps :
Life time (in hours): 1500 - 2000, 2000 - 2500, 2500 - 3000, 3000 - 3500, 3500 - 4000, 4000 - 4500, 4500 - 5000
Number of lamps: 14, 56, 60, 86, 74, 62, 48
Find the median life time of a lamp.v
SolutionCumulative frequencies are $14,70,130,216,290,352,400$. Here $n=400$, so $n/2=200$. The median class is $3000-3500$, with $l=3000$, $h=500$, $cf=130$, $f=86$. Median $=3000+\dfrac{200-130}{86}\times500=3406.98$ hours approximately.
Answer:The median lifetime is approximately $3406.98$ hours.
Q.6100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
Number of letters: 1 - 4, 4 - 7, 7 - 10, 10 - 13, 13 - 16, 16 - 19
Number of surnames: 6, 30, 40, 16, 4, 4
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.v
SolutionClass marks are $2.5,5.5,8.5,11.5,14.5,17.5$, so $\sum f_i x_i=832$ and mean $=\dfrac{832}{100}=8.32$. Cumulative frequencies are $6,36,76,92,96,100$; median class is $7-10$, so median $=7+\dfrac{50-36}{40}\times3=8.05$. The modal class is also $7-10$, so mode $=7+\dfrac{40-30}{2(40)-30-16}\times3=7.88$ approximately.
Answer:Median $=8.05$ letters, mean $=8.32$ letters, and mode $\approx7.88$ letters.
Q.7The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Weight (in kg): 40 - 45, 45 - 50, 50 - 55, 55 - 60, 60 - 65, 65 - 70, 70 - 75
Number of students: 2, 3, 8, 6, 6, 3, 2v
SolutionCumulative frequencies are $2,5,13,19,25,28,30$. Here $n=30$, so $n/2=15$. The median class is $55-60$, with $l=55$, $h=5$, $cf=13$, $f=6$. Median $=55+\dfrac{15-13}{6}\times5=56.67$ kg approximately.
Answer:The median weight is approximately $56.67$ kg.