# Ncert Solutions For Class 10 Maths Ex 14.3

## Ncert Solutions For Class 10 Maths Chapter 14 Ex 14.3

Question 1:

The following frequency distribution gives the monthly consumption of an electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.

 Monthly consumption(in units) No. of customers 65-85 4 85-105 5 105-125 13 125-145 20 145-165 14 165-185 8 185-205 4

Solution:

 Class Interval Frequency Cumulative frequency 65-85 4 4 85-105 5 9 105-125 13 22 125-145 20 42 145-165 14 56 165-185 8 64 185-205 4 68 N=68

Where, n = 68 and hence n2=34$\frac{n}{2}=34$

Hence, the median class is 125-145 with cumulative frequency = 42

Where, l = 125, n = 68, cf = 22, f = 20, h = 20

Median is calculated as follows:

Median=l+(n2cff)Xh$l+\left ( \frac{\frac{n}{2}-cf}{f} \right )Xh$

=125+(342220)X20$125+\left ( \frac{34-22}{20} \right )X20$

=125+12=137

Mode = Modal class = 125-145, f1=20,f0=13,f2=14$f_{1} = 20, f_{0} = 13, f_{2} = 14$ & h = 20

Mode=l+(f1f02Xf1f0f2)Xh$l+\left ( \frac{f_{1}-f_{0}}{2Xf_{1}-f_{0}-f_{2}} \right )Xh$

=125+(20132X201314)X20$125+\left ( \frac{20-13}{2X20-13-14} \right )X20$

=125+713X20$125+\frac{7}{13}X20$

=125+10.77

=135.77

Calculate the Mean:

 Class Interval fi xi di=xi-a ui=di/h fiui 65-85 4 75 -60 -3 -12 85-105 5 95 -40 -2 -10 105-125 13 115 -20 -1 -13 125-145 20 135 0 0 0 145-165 14 155 20 1 14 165-185 8 175 40 2 16 185-205 4 195 60 3 12 Sum fi= 68 Sum fiui = 7

x¯=a+fiuifiXh$\bar{x}=a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}X h$ =135+768X20$=135+\frac{7}{68}X20$

=137.05

Mean, median and mode are more/less equal in this distribution.

Question 2:

If the median of a distribution given below is 28.5 then, find the value of an x &y.

 Class Interval Frequency 0-10 5 10-20 x 20-30 20 30-40 15 40-50 y 50-60 5 Total 60

Solution:

n = 60

Where,n2$\frac{n}{2}$ = 30

Median class is 20 – 30 with a cumulative frequency = 25 + x

Lower limit of median class = 20, cf = 5 + x , f = 20 & h = 10

Median=l+(n2cff)Xh$\left ( \frac{\frac{n}{2}-cf}{f} \right )Xh$

Or,28.5=20+(305x20)X10$\left ( \frac{30-5-x}{20} \right )X10$

Or,25x2$\frac{25-x}{2}$

Or,   25-x=17

Or,    x= 25 – 17   =8

Now, from cumulative frequency, we can identify the value of x + y as follows:

60=5+20+15+5+x+y

Or, 45+x+y=60

Or, x + y=60-45=15

Hence, y = 15 – x= 15 – 8 = 7

Hence, x = 8  &  y = 7

Question 3:

The Life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to the  persons whose age is 18 years onwards but less than the 60 years.

 Age(in years) Number of policy holder Below 20 2 Below 25 6 Below 30 24 Below 35 45 Below 40 78 Below 45 89 Below 50 92 Below 55 98 Below 60 100

Solution:

 Class interval Frequency Cumulative frequency 15-20 2 2 20-25 4 6 25-30 18 24 30-35 21 45 35-40 33 78 40-45 11 89 45-50 3 92 50-55 6 98 55-60 2 100

Where, n = 100 and n2$\frac{n}{2}$ = 50

Median class = 35-45

Then, l = 35, cf = 45, f = 33 & h = 5

Median=l+(n2cff)Xh$\left ( \frac{\frac{n}{2}-cf}{f} \right )Xh$

=35+(504533)X5$=35+\left ( \frac{50-45}{33} \right )X5$ =35+(2533)$=35+\left ( \frac{25}{33} \right )$

=35.75

Question 4:

The lengths of 40 leaves in a plant are measured correctly to the nearest millimetre, and the data obtained is represented as in the following table:

 Length(in mm) Number of leaves 118-126 3 127-135 5 136-144 9 145-153 12 154-162 5 163-171 4 172-180 2

Find the median length of leaves.

Solution:

 Class Interval Frequency Cumulative frequency 117.5-126.5 3 3 126.5-135.5 5 8 135.5-144.5 9 17 144.5-153.5 12 29 153.5-162.5 5 34 162.5-171.5 4 38 171.5-180.5 2 40

Where, n = 40 and n2$\frac{n}{2}$ = 20

Median class = 144.5-153.5

then, l = 144.5, cf = 17, f = 12 & h = 9

Median=l+(n2cff)Xh$\left ( \frac{\frac{n}{2}-cf}{f} \right )Xh$

=144.5+(201712)X9$=144.5+\left ( \frac{20-17}{12} \right )X9$ =144.5+(94)$=144.5+\left ( \frac{9}{4} \right )$

=146.75

Question 5:

The following table gives the distribution of a life time of 400 neon lamps.

 Lifetime(in hours) Number of lamps 1500-2000 14 2000-2500 56 2500-3000 60 3000-3500 86 3500-4000 74 4000-4500 62 4500-5000 48

Find the median lifetime of a lamp.

Solution:

 Class Interval Frequency Cumulative 1500-2000 14 14 2000-2500 56 70 2500-3000 60 130 3000-3500 86 216 3500-4000 74 290 4000-4500 62 352 4500-5000 48 400

Where, n = 400 &n2$\frac{n}{2}$ = 200

Median class = 3000 – 3500

Therefore, l = 3000, cf = 130, f = 86 & h = 500

Median=l+(n2cff)Xh$\left ( \frac{\frac{n}{2}-cf}{f} \right )Xh$

=3000+(20013086)X500$=3000+\left ( \frac{200-130}{86} \right )X500$

=3000+406.97

=3406.97

Question 6:

In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in 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 number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.

Solution: To Calculate median:

 Class Interval Frequency Cumulative Frequency 1-4 6 6 4-7 30 36 7-10 40 76 10-13 16 92 13-16 4 96 16-19 4 100

Where, n = 100 &n2$\frac{n}{2}$ = 50

Median class = 7-10

Therefore, l = 7, cf = 36, f = 40 & h = 3

Median=l+(n2cff)Xh$\left ( \frac{\frac{n}{2}-cf}{f} \right )Xh$

=7+(503640)X3$=7+\left ( \frac{50-36}{40} \right )X3$ =7+(1440)X3$=7+\left ( \frac{14}{40} \right )X3$

=8.05

Calculate the Mode:

Modal class = 7-10,

Where, l = 7, f1 = 40, f0 = 30, f2 = 16 & h = 3

Mode=l+(f1f02Xf1f0f2)Xh$l+\left ( \frac{f_{1}-f_{0}}{2Xf_{1}-f_{0}-f_{2}} \right )Xh$

=7+(40302X403016)X3$7+\left ( \frac{40-30}{2X40-30-16} \right )X3$

=7+1034X3$7+\frac{10}{34}X3$

=7.88

Calculate the Mean:

 Class Interval fi xi fixi 1-4 6 2.5 15 4-7 30 5.5 165 7-10 40 8.5 340 10-13 16 11.5 184 13-16 4 14.5 51 16-19 4 17.5 70 Sum fi = 100 Sum fixi = 825

x¯=a+fiuifiXh$\bar{x}=a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}X h$ =825100$=\frac{825}{100}$

=8.25

Q7. The distributions of below gives a weight of 30 students of a class. Find the median weight of a 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 2

Solution:

 Class Interval Frequency Cumulative frequency 40-45 2 2 45-50 3 5 50-55 8 13 55-60 6 19 60-65 6 25 65-70 3 28 70-75 2 30

Where, n = 30 and n2$\frac{n}{2}$= 15

median class = 55-60

therefore, l = 55, cf = 13, f = 6 & h = 5

Median=l+(n2cff)Xh$\left ( \frac{\frac{n}{2}-cf}{f} \right )Xh$

=55+(15136)X5$=55+\left ( \frac{15-13}{6} \right )X5$ =55+(53)$=55+\left ( \frac{5}{3} \right )$

=56.67

Question 7:

The distribution shows the pocket money of students in an area. The mean pocket money is Rs 18. Find the missing frequency f.

 Daily pocket allowance (in Rs) 11-13 13-15 15-17 17-19 19-21 21-23 23-25 Number of children 07 06 09 013 F 05 04

Solution:

 Class interval fi xi fixi 11-13 07 012 84 13-15 06 014 84 15-17 09 016 144 17-19 013 018 234 19-21 f 020 20f 21-23 05 022 110 23-25 04 024 96 Sum fi = 44+f Sum fixi = 752+20f

We have,

Mean pocket money, xi$x_{i}$ = 18(given)

Hence substituting for fixi,

18(44+f) = 752 + 20f

Or, 792 + 18f = 752 + 20f

Or, 2f = 40

f=20$\rightarrow f=20$