Statistics — CBSE Class 10 Maths Textbook Solutions

Since the class marks $x_i$x_i and frequencies $f_i$f_i here are small numbers, the direct method is the most convenient choice.

Using the direct method:
$\bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{162}{20}=8.1$x= f_ix_i/ f_i=162/20=8.1

The mean number of plants per house is 8.1. The direct method was used because both the class marks and the frequencies are small, so the products $f_ix_i$f_ix_i are easy to compute directly without shifting the origin.

Since the class marks are large numbers, the step-deviation method is used, taking assumed mean $a=550$a=550 and class size $h=20$h=20.

$\bar{x}=a+\left(\frac{\sum f_iu_i}{\sum f_i}\right)\times h=550+\left(\frac{-12}{50}\right)\times 20=550-4.8=545.2$x=a+( f_iu_i/ f_i)× h=550+(-12/50)× 20=550-4.8=545.2

The mean daily wage of the workers is Rs 545.20.

Use the assumed mean method with $a=18$a=18 (the class mark of 17-19) and $h=2$h=2.

Since the mean is given as 18 (which equals the assumed mean $a$a), we need:
$\bar{x}=a+\left(\frac{\sum f_iu_i}{\sum f_i}\right)\times h \implies 18=18+\left(\frac{f-20}{44+f}\right)\times 2$x=a+( f_iu_i/ f_i)× h 18=18+(f-20/44+f)× 2
This gives $\dfrac{f-20}{44+f}=0$f-20/44+f=0, so $f-20=0$f-20=0.
$f=20$f=20
The missing frequency is f = 20.

Use the step-deviation method with $a=75.5$a=75.5 (class mark of 74-77) and $h=3$h=3.

$\bar{x}=a+\left(\frac{\sum f_iu_i}{\sum f_i}\right)\times h=75.5+\left(\frac{4}{30}\right)\times 3=75.5+0.4=75.9$x=a+( f_iu_i/ f_i)× h=75.5+(4/30)× 3=75.5+0.4=75.9

The mean number of heartbeats per minute is 75.9.

The classes given are inclusive (50-52, 53-55, ...), but since the class marks are unaffected by this, we can work with them directly. Since the frequencies are large numbers, the step-deviation method is the most convenient, with $a=57$a=57 and $h=3$h=3.

$\bar{x}=a+\left(\frac{\sum f_iu_i}{\sum f_i}\right)\times h=57+\left(\frac{25}{400}\right)\times 3=57+0.1875=57.1875$x=a+( f_iu_i/ f_i)× h=57+(25/400)× 3=57+0.1875=57.1875

The mean number of mangoes per box is approximately 57.19. The step-deviation method was chosen because the frequencies (number of boxes) are large numbers, making direct multiplication tedious.

Use the step-deviation method with $a=225$a=225 (class mark of 200-250) and $h=50$h=50.

$\bar{x}=a+\left(\frac{\sum f_iu_i}{\sum f_i}\right)\times h=225+\left(\frac{-7}{25}\right)\times 50=225-14=211$x=a+( f_iu_i/ f_i)× h=225+(-7/25)× 50=225-14=211

The mean daily expenditure on food is Rs 211.

Since the class marks are small decimal numbers, the direct method is convenient.

$\bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{2.96}{30}=0.09867$x= f_ix_i/ f_i=2.96/30=0.09867

The mean concentration of $SO_2$SO_2 in the air is approximately 0.099 ppm.

The class sizes here are unequal, so the step-deviation method cannot be applied directly. We use the direct method.

$\bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{499}{40}=12.475$x= f_ix_i/ f_i=499/40=12.475

The mean number of days a student was absent is approximately 12.48 days.

Use the step-deviation method with $a=70$a=70 (class mark of 65-75) and $h=10$h=10.

$\bar{x}=a+\left(\frac{\sum f_iu_i}{\sum f_i}\right)\times h=70+\left(\frac{-2}{35}\right)\times 10=70-0.5714=69.4286$x=a+( f_iu_i/ f_i)× h=70+(-2/35)× 10=70-0.5714=69.4286

The mean literacy rate is approximately 69.43%.

Mode: The maximum frequency is 23, so the modal class is 35-45.
Here $l=35$l=35, $h=10$h=10, $f_1=23$f_1=23, $f_0=21$f_0=21, $f_2=14$f_2=14.
$\text{Mode}=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h=35+\left(\frac{23-21}{46-21-14}\right)\times 10=35+\left(\frac{2}{11}\right)\times 10=35+1.818=36.8$Mode=l+(f_1-f_0/2f_1-f_0-f_2)× h=35+(23-21/46-21-14)× 10=35+(2/11)× 10=35+1.818=36.8

Mean: Use the step-deviation method with $a=40$a=40 (class mark of 35-45) and $h=10$h=10.

$\bar{x}=40+\left(\frac{-37}{80}\right)\times 10=40-4.625=35.375$x=40+(-37/80)× 10=40-4.625=35.375

Mode = 36.8 years, Mean = 35.37 years. The maximum number of patients admitted to the hospital are of age 36.8 years (approximately), while on average, the age of a patient admitted is 35.37 years.

The maximum frequency is 61, so the modal class is 60-80.
Here $l=60$l=60, $h=20$h=20, $f_1=61$f_1=61, $f_0=52$f_0=52, $f_2=38$f_2=38.
$\text{Mode}=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h=60+\left(\frac{61-52}{122-52-38}\right)\times 20=60+\left(\frac{9}{32}\right)\times 20=60+5.625=65.625$Mode=l+(f_1-f_0/2f_1-f_0-f_2)× h=60+(61-52/122-52-38)× 20=60+(9/32)× 20=60+5.625=65.625

The modal lifetime of the components is 65.625 hours.

Mode: The maximum frequency is 40, so the modal class is 1500-2000.
Here $l=1500$l=1500, $h=500$h=500, $f_1=40$f_1=40, $f_0=24$f_0=24, $f_2=33$f_2=33.
$\text{Mode}=1500+\left(\frac{40-24}{80-24-33}\right)\times 500=1500+\left(\frac{16}{23}\right)\times 500=1500+347.83=1847.83$Mode=1500+(40-24/80-24-33)× 500=1500+(16/23)× 500=1500+347.83=1847.83

Mean: Use the step-deviation method with $a=3250$a=3250 (class mark of 3000-3500) and $h=500$h=500.

$\bar{x}=3250+\left(\frac{-235}{200}\right)\times 500=3250-587.5=2662.5$x=3250+(-235/200)× 500=3250-587.5=2662.5

Modal monthly expenditure = Rs 1847.83, Mean monthly expenditure = Rs 2662.5.

Mode: The maximum frequency is 10, so the modal class is 30-35.
Here $l=30$l=30, $h=5$h=5, $f_1=10$f_1=10, $f_0=9$f_0=9, $f_2=3$f_2=3.
$\text{Mode}=30+\left(\frac{10-9}{20-9-3}\right)\times 5=30+\left(\frac{1}{8}\right)\times 5=30+0.625=30.625$Mode=30+(10-9/20-9-3)× 5=30+(1/8)× 5=30+0.625=30.625

Mean: Use the step-deviation method with $a=32.5$a=32.5 (class mark of 30-35) and $h=5$h=5.

$\bar{x}=32.5+\left(\frac{-23}{35}\right)\times 5=32.5-3.2857=29.214$x=32.5+(-23/35)× 5=32.5-3.2857=29.214

Mode = 30.6, Mean = 29.2. Most states/U.T. have a student-teacher ratio of about 30.6, while the average student-teacher ratio across all states/U.T. is 29.2.

The maximum frequency is 18, so the modal class is 4000-5000.
Here $l=4000$l=4000, $h=1000$h=1000, $f_1=18$f_1=18, $f_0=4$f_0=4, $f_2=9$f_2=9.
$\text{Mode}=4000+\left(\frac{18-4}{36-4-9}\right)\times 1000=4000+\left(\frac{14}{23}\right)\times 1000=4000+608.7=4608.7$Mode=4000+(18-4/36-4-9)× 1000=4000+(14/23)× 1000=4000+608.7=4608.7

The mode of the data is 4608.7 runs.

The maximum frequency is 20, so the modal class is 40-50.
Here $l=40$l=40, $h=10$h=10, $f_1=20$f_1=20, $f_0=12$f_0=12, $f_2=11$f_2=11.
$\text{Mode}=40+\left(\frac{20-12}{40-12-11}\right)\times 10=40+\left(\frac{8}{17}\right)\times 10=40+4.706=44.706$Mode=40+(20-12/40-12-11)× 10=40+(8/17)× 10=40+4.706=44.706

The mode of the data is 44.7 cars.

Median: Build the cumulative frequency column.

$n=68$n=68, so $\frac{n}{2}=34$n/2=34. The cf just greater than 34 is 42, so the median class is 125-145.
$l=125$l=125, $cf=22$cf=22 (cf of preceding class), $f=20$f=20, $h=20$h=20.
$\text{Median}=125+\left(\frac{34-22}{20}\right)\times 20=125+12=137$Median=125+(34-22/20)× 20=125+12=137

Mean: Use the step-deviation method, $a=135$a=135 (class mark of 125-145), $h=20$h=20.

$\bar{x}=135+\left(\frac{7}{68}\right)\times 20=135+2.059=137.06 \approx 137.05$x=135+(7/68)× 20=135+2.059=137.06 137.05

Mode: The maximum frequency is 20, so the modal class is 125-145.
$l=125$l=125, $h=20$h=20, $f_1=20$f_1=20, $f_0=13$f_0=13, $f_2=14$f_2=14.
$\text{Mode}=125+\left(\frac{20-13}{40-13-14}\right)\times 20=125+\left(\frac{7}{13}\right)\times 20=125+10.77=135.76$Mode=125+(20-13/40-13-14)× 20=125+(7/13)× 20=125+10.77=135.76

Median = 137 units, Mean = 137.05 units, Mode = 135.76 units. The three measures are approximately equal here, which shows this is a fairly symmetric (not badly skewed) distribution.

Total frequency: $5+x+20+15+y+5=60 \implies x+y=15$5+x+20+15+y+5=60 x+y=15 ... (1)

Cumulative frequencies: 0-10: 5; 10-20: $5+x$5+x; 20-30: $25+x$25+x; 30-40: $40+x$40+x; 40-50: $40+x+y$40+x+y; 50-60: $45+x+y$45+x+y.

$n=60$n=60, so $\frac{n}{2}=30$n/2=30. The median is 28.5, which lies in the class 20-30 (since $25+x$25+x will be the cf just past 30 for reasonable $x$x). So the median class is 20-30.
$l=20$l=20, $cf=5+x$cf=5+x, $f=20$f=20, $h=10$h=10.
$\text{Median}=l+\left(\frac{\frac{n}{2}-cf}{f}\right)\times h$Median=l+(n/2-cff)× h
$28.5=20+\left(\frac{30-(5+x)}{20}\right)\times 10$28.5=20+(30-(5+x)/20)× 10
$28.5-20=\frac{(25-x)\times 10}{20}$28.5-20=(25-x)× 10/20
$8.5=\frac{25-x}{2}$8.5=25-x/2
$17=25-x \implies x=8$17=25-x x=8

Substituting into (1): $8+y=15 \implies y=7$8+y=15 y=7.

x = 8, y = 7.

This is a cumulative frequency (less than type) table. Convert it to a class-wise frequency table by taking successive differences, with classes 18-20, 20-25, 25-30, ..., 55-60.

$n=100$n=100, so $\frac{n}{2}=50$n/2=50. The cf just greater than 50 is 78, so the median class is 35-40.
$l=35$l=35, $cf=45$cf=45 (cf of preceding class), $f=33$f=33, $h=5$h=5.
$\text{Median}=35+\left(\frac{50-45}{33}\right)\times 5=35+\left(\frac{5}{33}\right)\times 5=35+0.7576=35.76$Median=35+(50-45/33)× 5=35+(5/33)× 5=35+0.7576=35.76

The median age of the policy holders is 35.76 years.

Convert to continuous classes (subtract 0.5 from lower limits, add 0.5 to upper limits) and build the cumulative frequency column.

$n=40$n=40, so $\frac{n}{2}=20$n/2=20. The cf just greater than 20 is 29, so the median class is 144.5-153.5.
$l=144.5$l=144.5, $cf=17$cf=17, $f=12$f=12, $h=9$h=9.
$\text{Median}=144.5+\left(\frac{20-17}{12}\right)\times 9=144.5+\left(\frac{3}{12}\right)\times 9=144.5+2.25=146.75$Median=144.5+(20-17/12)× 9=144.5+(3/12)× 9=144.5+2.25=146.75

The median length of the leaves is 146.75 mm.

Build the cumulative frequency column.

$n=400$n=400, so $\frac{n}{2}=200$n/2=200. The cf just greater than 200 is 216, so the median class is 3000-3500.
$l=3000$l=3000, $cf=130$cf=130, $f=86$f=86, $h=500$h=500.
$\text{Median}=3000+\left(\frac{200-130}{86}\right)\times 500=3000+\left(\frac{70}{86}\right)\times 500=3000+406.98=3406.98$Median=3000+(200-130/86)× 500=3000+(70/86)× 500=3000+406.98=3406.98

The median life time of a lamp is 3406.98 hours.

Median: Build the cumulative frequency column.

$n=100$n=100, so $\frac{n}{2}=50$n/2=50. The cf just greater than 50 is 76, so the median class is 7-10.
$l=7$l=7, $cf=36$cf=36, $f=40$f=40, $h=3$h=3.
$\text{Median}=7+\left(\frac{50-36}{40}\right)\times 3=7+\left(\frac{14}{40}\right)\times 3=7+1.05=8.05$Median=7+(50-36/40)× 3=7+(14/40)× 3=7+1.05=8.05

Mean: Use the step-deviation method, $a=8.5$a=8.5 (class mark of 7-10), $h=3$h=3.

$\bar{x}=8.5+\left(\frac{-6}{100}\right)\times 3=8.5-0.18=8.32$x=8.5+(-6/100)× 3=8.5-0.18=8.32

Mode: The maximum frequency is 40, so the modal class is 7-10.
$l=7$l=7, $h=3$h=3, $f_1=40$f_1=40, $f_0=30$f_0=30, $f_2=16$f_2=16.
$\text{Mode}=7+\left(\frac{40-30}{80-30-16}\right)\times 3=7+\left(\frac{10}{34}\right)\times 3=7+0.882=7.88$Mode=7+(40-30/80-30-16)× 3=7+(10/34)× 3=7+0.882=7.88

Median = 8.05, Mean = 8.32, Modal size = 7.88.

Build the cumulative frequency column.

$n=30$n=30, so $\frac{n}{2}=15$n/2=15. The cf just greater than 15 is 19, so the median class is 55-60.
$l=55$l=55, $cf=13$cf=13, $f=6$f=6, $h=5$h=5.
$\text{Median}=55+\left(\frac{15-13}{6}\right)\times 5=55+\left(\frac{2}{6}\right)\times 5=55+1.6667=56.67$Median=55+(15-13/6)× 5=55+(2/6)× 5=55+1.6667=56.67

The median weight of the students is 56.67 kg.