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Class 10 NCERT Solutions- Chapter 14 Statistics – Exercise 14.3

Last Updated : 03 Mar, 2021
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Question 1. The following frequency distribution gives the monthly consumption of 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
65-854
85-1055
105-12513
125-14520
145-16514
165-1858
185-2054

Solution:

 Total number of consumer n = 68 

n/2 =34

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

Here, l = 125, n = 68, Cf = 22, f = 20, h = 20

Now we find the median:

Median = l +\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h  

=125+\frac{(34−22)}{20} × 20

= 125 + 12 = 137

Hence, the median is 137

Now we find the mode:

Modal class = 125 – 145,

Frequencies are

f1 = 20, f0 = 13, f2 = 14 & h = 20

Mode l+ \left[\frac{(f1-f0)}{(2f1-f0-f2)}\right]×h

On substituting the values in the given formula, we get

Mode = 125 + \frac{(20-13)}{(40-13-14)}×20

= 125 + 140/13 

= 125 + 10.77

= 135.77

Hence, the mode is 135.77

Now we find the mean:

Class Intervalfixidi = xi – aui = di/hfiui
65-85475-60-3-12
85-105595-40-2-10
105-12513115-20-1-13
125-14520135000
145-1651415520114
165-185817540216
185-205419560312
 Sum fi = 68   Sum fiui = 7

\bar{x} =a+h \frac{∑f_iu_i}{∑f_i}

= 135 + 20(7/68)

= 137.05

Hence, the mean is 137.05

Now, on comparing the median, mean, and mode, we found that 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 x & y.

Class IntervalFrequency
0-105
10-20x
20-3020
30-4015
40-50y
50-605
Total60

Solution:

According to the question

The total number of observations are n = 60

Median of the given data = 28.5

n/2 = 30  

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

Lower limit of median class, l = 20,

Cf = 5 + x,

f = 20 & h = 10

Now we find the median:

Median = l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

28.5 = 20+\frac{(30−5−x)}{20} × 10

8.5 = (25 – x)/2

17 = 25 – x

Therefore, x = 8

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

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

On substituting the values of x, we will find the value of y

60 = 5 + 20 + 15 + 5 + 8 + y

y = 60 – 53

y = 7

So the value of a is 8 and y is 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 202
Below 256
Below 3024
Below 3545
Below 4078
Below 4589
Below 5092
Below 5598
Below 60100

Solution: 

According to the given question the table is 

Class intervalFrequencyCumulative frequency
15-2022
20-2546
25-301824
30-352145
35-403378
40-451189
45-50392
50-55698
55-602100

Given data: n = 100 and n/2 = 50

Median class = 35 – 45

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

Now we find the median:

Median = l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

Median = 35+\frac{(50-45)}{33} × 5

= 35 + 5(5/33) 

= 35.75

Hence, the median age is 35.75 years.

Question 4. The lengths of 40 leaves in a plant are measured correctly to the nearest millimeter, and the data obtained is represented as in the following table:

Length (in mm)Number of leaves
118-1263
127-1355
136-1449
145-15312
154-1625
163-1714
172-1802

Find the median length of leaves.             

Solution:

The data in the given table are not continuous to reduce 0.5 in the lower limit and add 0.5 in the upper limit.

We get a new table:

Class IntervalFrequencyCumulative frequency
117.5-126.533
126.5-135.558
135.5-144.5917
144.5-153.51229
153.5-162.5534
162.5-171.5438
171.5-180.5240

From the given table

n = 40 and n/2 = 20

Median class = 144.5 – 153.5

l = 144.5,

cf = 17, f = 12 & h = 9

Now we find the median:

Median = l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

Median = 144.5+\frac{(20-17)}{12}×9

= 144.5 + 9/4 

= 146.75 mm

Hence, the median length of the leaves is 146.75 mm.

Question 5. The following table gives the distribution of a life time of 400 neon lamps.

Lifetime (in hours)Number of lamps
1500-200014
2000-250056
2500-300060
3000-350086
3500-400074
4000-450062
4500-500048

Find the median lifetime of a lamp.

Solution:

According to the question

Class IntervalFrequencyCumulative
1500-20001414
2000-25005670
2500-300060130
3000-350086216
3500-400074290
4000-450062352
4500-500048400

n = 400 and n/2 = 200

Median class = 3000 – 3500

l = 3000, Cf = 130,

f = 86 & h = 500

Now we find the median:

Median = l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

Median = 3000 + \frac{(200-130)}{86} × 500

= 3000 + 35000/86 = 3000 + 406.97

= 3406.97

Hence, the median lifetime of the lamps is 3406.97 hours

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 letters1-44-77-1010-1313-1616-19
Number of surnames630401644

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:

According to the question

Class IntervalFrequencyCumulative Frequency
1-466
4-73036
7-104076
10-131692
13-16496
16-194100

n = 100 and n/2 = 50

Median class = 7 – 10

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

Now we find the median:

Median = l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

Median = 7+\frac{(50-36)}{40} × 3

Median = 7 + 42/40 = 8.05

Hence, the median is 8.05

Now we find the mode:

Modal class = 7 – 10,

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

Mode = l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h

On substituting the values in the given formula, we get

Mode = 7+\frac{(40-30)}{(2×40-30-16)} × 3

= 7 + 30/34 = 7.88

Hence, the mode is 7.88

Now we find the mean:

Class Intervalfixifixi
1-462.515
4-7305.5165
7-10408.5340
10-131611.5184
13-16414.551
16-19417.570
 Sum fi = 100 Sum fixi = 825

Mean = \bar{x}= \frac{∑f_i x_i }{∑f_i}

= 825/100 = 8.25

Hence, the mean is 8.25

Question 7. The distributions of below give a weight of 30 students of a class. Find the median weight of a student.

Weight(in kg)40-4545-5050-5555-6060-6565-7070-75
Number of students2386632

Solution:

According to the question

Class IntervalFrequencyCumulative frequency
40-4522
45-5035
50-55813
55-60619
60-65625
65-70328
70-75230

n = 30 and n/2 = 15

Median class = 55 – 60

l = 55, Cf = 13, f = 6 & h = 5

Now we find the median:

Median = l +\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

Median = 55+\frac{(15-13)}{6}×5

= 55 + 10/6 = 55 + 1.666

= 56.67

Hence, the median weight of the students is 56.67



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