CBSE · NCERT · Class 9 Maths · Chapter 12

NCERT Solutions: Class 9 Maths Chapter 12 - Statistics

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Chapter-wise NCERT intext questions and exercise answers for Statistics, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
Sections in this chapter
Exercise 12.1 6Exercise 12.2 6Exercise 12.3 6
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1Exercise 12.16 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
Solution

The class marks are $1,3,5,7,9,11,13$. Compute $\sum f_ix_i=1(1)+2(3)+1(5)+5(7)+6(9)+2(11)+3(13)=162$, and $\sum f_i=20$. Hence mean $\bar{x}=\dfrac{162}{20}=8.1$.

Answer:

The mean number of plants per house is $8.1$ plants. The direct method is convenient because the class marks and frequencies are small.

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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, 600-620, 620-640. Number of workers: 12, 14, 8, 6, 4, 3, 3. Find the mean daily wages of the workers of the factory by using an appropriate method.v
Solution

Use class marks $510,530,550,570,590,610,630$. Then $\sum f_ix_i=12(510)+14(530)+8(550)+6(570)+4(590)+3(610)+3(630)=27260$, and $\sum f_i=50$. Therefore $\bar{x}=\dfrac{27260}{50}=545.2$.

Answer:

The mean daily wage is $₹545.20$.

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Q.3The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹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, 4.v
Solution

The class marks are $12,14,16,18,20,22,24$. Total frequency $=44+f$. Also $\sum f_ix_i=7(12)+6(14)+9(16)+13(18)+f(20)+5(22)+4(24)=752+20f$. Given mean $18$, so $\dfrac{752+20f}{44+f}=18$. Hence $752+20f=792+18f$, giving $2f=40$ and $f=20$.

Answer:

$f=20$.

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Q.4Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method. Number of heart beats 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, 2.v
Solution

The class marks are $66.5,69.5,72.5,75.5,78.5,81.5,84.5$. Since the intervals are equal, take assumed mean $a=75.5$ and class width $h=3$. The deviations $u_i=(x_i-a)/h$ are $-3,-2,-1,0,1,2,3$. Then $\sum f_iu_i=2(-3)+4(-2)+3(-1)+8(0)+7(1)+4(2)+2(3)=4$, and $\sum f_i=30$. Mean $=a+h\dfrac{\sum f_iu_i}{\sum f_i}=75.5+3\cdot\dfrac4{30}=75.9$.

Answer:

The mean is $75.9$ heart beats per minute.

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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. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose? Number of mangoes: 50-52, 53-55, 56-58, 59-61, 62-64. Number of boxes: 15, 110, 135, 115, 25.v
Solution

The classes are inclusive, but their class marks are $51,54,57,60,63$. Using the assumed mean method with $a=57$, compute deviations $d_i=x_i-a=-6,-3,0,3,6$. Then $\sum f_id_i=15(-6)+110(-3)+135(0)+115(3)+25(6)=75$, and $\sum f_i=400$. Hence mean $=57+\dfrac{75}{400}=57.1875\approx57.19$.

Answer:

The mean number of mangoes per box is $57.19$.

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Q.6The 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. Daily expenditure (in ₹): 100-150, 150-200, 200-250, 250-300, 300-350. Number of households: 4, 5, 12, 2, 2.v
Solution

Class marks are $125,175,225,275,325$. Compute $\sum f_ix_i=4(125)+5(175)+12(225)+2(275)+2(325)=5275$, and $\sum f_i=25$. Hence $\bar{x}=\dfrac{5275}{25}=211$.

Answer:

The mean daily expenditure is $₹211$.

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2Exercise 12.26 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
Solution

The modal class is $35-45$ because it has the highest frequency $23$. With $l=35$, $h=10$, $f_1=23$, $f_0=21$, $f_2=14$, mode $=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}h=35+\dfrac{2}{46-21-14}\times10=35+\dfrac{20}{11}\approx36.8$. For the mean, class marks are $10,20,30,40,50,60$, so $\sum f_ix_i=2830$ and $\sum f_i=80$. Mean $=\dfrac{2830}{80}=35.375$.

Answer:

Mode $=36.8$ years and mean $=35.375$ years. The typical patient age is around $36$ years.

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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
Solution

The modal class is $60-80$. Here $l=60$, $h=20$, $f_1=61$, $f_0=52$, and $f_2=38$. Mode $=60+\dfrac{61-52}{2(61)-52-38}\times20=60+\dfrac{9}{32}\times20=65.625$ hours, approximately $65.6$ hours.

Answer:

The modal lifetime is approximately $65.6$ hours.

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Q.3The following data gives the distribution of total monthly household expenditure of 200 families of a village. 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, 7. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.v
Solution

The modal class is $1500-2000$. Here $l=1500$, $h=500$, $f_1=40$, $f_0=24$, $f_2=33$. Mode $=1500+\dfrac{40-24}{2(40)-24-33}\times500=1500+\dfrac{16}{23}\times500\approx1847.83$. For the mean, class marks are $1250,1750,2250,2750,3250,3750,4250,4750$. Then $\sum f_ix_i=532500$ and $\sum f_i=200$, so mean $=\dfrac{532500}{200}=₹2662.50$.

Answer:

Modal expenditure $=₹1847.83$ approximately; mean expenditure $=₹2662.50$.

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Q.4The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. 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, 2. Find the mode and mean of this data. Interpret the two measures.v
Solution

The 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, class marks are $17.5,22.5,27.5,32.5,37.5,42.5,47.5,52.5$. Then $\sum f_ix_i=1022$ and $\sum f_i=35$, so mean $=\dfrac{1022}{35}=29.2$. The modal ratio says the most common class is around $31$ students per teacher; the mean gives the overall average ratio.

Answer:

Mode $=30.625$ students per teacher and mean $=29.2$ students per teacher.

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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
Solution

The modal class is $4000-5000$. Here $l=4000$, $h=1000$, $f_1=18$, $f_0=4$, and $f_2=9$. Mode $=4000+\dfrac{18-4}{2(18)-4-9}\times1000=4000+\dfrac{14}{23}\times1000\approx4608.7$.

Answer:

The modal number of runs is approximately $4608.7$.

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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. 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, 8. Find the mode of the data.v
Solution

The modal class is $40-50$. With $l=40$, $h=10$, $f_1=20$, $f_0=12$, and $f_2=11$, mode $=40+\dfrac{20-12}{2(20)-12-11}\times10=40+\dfrac{8}{17}\times10\approx44.7$.

Answer:

The mode is approximately $44.7$ cars.

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3Exercise 12.36 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, 4.v
Solution

For the median, $N=68$, so $N/2=34$. Cumulative frequencies are $4,9,22,42,56,64,68$, so the median class is $125-145$. Here $l=125$, $h=20$, $f=20$, and preceding cumulative frequency $cf=22$. Median $=125+\dfrac{34-22}{20}\times20=137$. For the mean, class marks are $75,95,115,135,155,175,195$ and $\sum f_ix_i=9320$, so mean $=\dfrac{9320}{68}\approx137.06$. For the mode, modal class is $125-145$, and mode $=125+\dfrac{20-13}{40-13-14}\times20=135.77$ units approximately.

Answer:

Median $=137$ units, mean $=137.06$ units approximately, and mode $=135.77$ units approximately. The three measures are close, so consumption is centred near $136$-$137$ units.

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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 frequency: 60.v
Solution

Since the total frequency is $60$, $5+x+20+15+y+5=60$, so $x+y=15$. The median $28.5$ lies in class $20-30$. Here $l=20$, $h=10$, $f=20$, and preceding cumulative frequency $cf=5+x$. Using $28.5=20+\dfrac{30-(5+x)}{20}\times10$, we get $8.5=\dfrac{25-x}{2}$, so $x=8$. Hence $y=15-8=7$.

Answer:

$x=8$ and $y=7$.

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Q.3A life insurance agent found the following data about 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, 100.v
Solution

Convert 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 cumulative frequency before $35-40$ is $45$, so the median class is $35-40$. Thus median $=35+\dfrac{50-45}{33}\times5=35+\dfrac{25}{33}\approx35.76$ years.

Answer:

The median age is approximately $35.76$ years.

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

Because measurements are correct to one millimetre, use continuous class boundaries: $117.5-126.5$, $126.5-135.5$, $135.5-144.5$, $144.5-153.5$, etc. Here $N=40$, so $N/2=20$. Cumulative frequencies are $3,8,17,29,34,38,40$, so the median class is $144.5-153.5$. With $l=144.5$, $h=9$, $f=12$, and $cf=17$, median $=144.5+\dfrac{20-17}{12}\times9=146.75$ mm.

Answer:

The median length is approximately $146.75$ mm.

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Q.5The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp. 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.v
Solution

Here $N=400$, so $N/2=200$. Cumulative frequencies are $14,70,130,216,290,352,400$, so the median class is $3000-3500$. With $l=3000$, $h=500$, $f=86$, and $cf=130$, median $=3000+\dfrac{200-130}{86}\times500\approx3406.98$ hours.

Answer:

The median lifetime is approximately $3406.98$ hours.

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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. Determine the median number of letters in the surnames. Number of letters: 1-4, 4-7, 7-10, 10-13, 13-16, 16-19. Number of surnames: 6, 30, 40, 16, 4, 4.v
Solution

Here $N=100$, so $N/2=50$. Cumulative frequencies are $6,36,76,92,96,100$, so the median class is $7-10$. With $l=7$, $h=3$, $f=40$, and $cf=36$, median $=7+\dfrac{50-36}{40}\times3=8.05$.

Answer:

The median number of letters is $8.05$.

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Class 9 Maths