Maths Chapter-14: Statistics

Exercise – 1

Question 1:

A group of pupil has conducted a survey as a as the part of their college program, They collected examined data’s of plants from 20 houses and nearby areas. Find the mean of the collected plants from the given houses.

No. of Plants Taken from 20 houses	0-2	2-4	4-6	6-8	8-10	10-12	12-14
Given number of houses	2	1	5	1	2	6	3

Name the method you use for finding the mean and why?

Solution:

Class Interval	fi	xi	fi xi
0-2	2	1	1
2-4	1	3	6
4-6	5	5	5
6-8	1	7	35
8-10	2	9	54
10-12	6	11	22
12-14	3	13	40
	Sum fi=20		Sum fixi=163

Mean can be calculated as follows:

\(\bar{x}= \frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}=\frac{163}{20}=8.15\)

In this issues the \( f_{i} and x_{i}\) are small so the direct method is given to solve the problem.

 

Question2:

The distributed wages is given for 50 workers working in the factory.

Daily wages given to workers (in Rs)	100-120	120-140	140-160	160-180	180-200
Number of wage workers in factory	14	12	6	8	10

 

Find the mean of workers of daily wages given by the workers.

Solution:

In this case, value of xi is quite large and hence we should select the assumed mean method.

Class given Interval	fi	xi	di = xi – a	fi di
100-120	14	110	-40	-480
120-140	12	130	-20	-280
140-160	6	150	0	0
160-180	8	170	20	120
180-200	10	190	40	400
	Sum fi=50			Sum fi di = -240

 

Now, mean of deviations of daily wagesis calculated as follows:

\(\bar{d}=\frac{\sum f_{i}d_{i}}{\sum f_{i}}=\frac{-240}{50}\)

x = d + a = 150+(-4.8)= 145.20

 

Question 3:

This distribution shows that the money daily pocket given to students in the given area. The mean pocket money is Rs. 18. Find the f which is the missing frequency.

Daily money allowance (in Rs)	11-13	13-15	15-17	17-19	19-21	21-23	23-25
Number of children	7	6	9	13	f	5	4

 

Solution:

Class interval	fi	xi	fixi
11-13	7	12	84
13-15	6	14	84
15-17	13	16	144
17-19	9	18	234
19-21	f	20	20f
21-23	4	22	110
23-25	5	24	96
	Sum fi=44+f		Sum fixi= 752+20f

We have;

\(\bar{x}= \frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}\) \(18= \frac{752+20f}{44+ f_{i}}\)

18(44+18f) = 752+20f

792+18f = 752+20f

2f = 40

Missing frequency f= 20

 

Question4:

The number of 30 women were checked by one doctor and the distributed the heartbeat of them in following distribution table. Find the mean of heart beats of the given per minute for these thirty women, by choosing a suitable following method.

Heartbeat per minute in given number	65-68	68-71	71-74	74-77	77-80	80-83	83-86
Number of 30 women	2	3	4	7	8	2	4

Solution:

Class Interval	fi	xi	di = xi – a	fi di
65-68	2	66.5	-9	-18
68-71	3	69.5	-6	-24
71-74	4	72.5	-3	-9
74-77	7	75.5	0	0
77-80	8	78.5	3	21
80-83	2	81.5	6	24
83-86	4	84.5	9	18
	Sum fi= 30			Sum fi di = 12

Now, mean can be calculated as follows:

\(\bar{d}=\frac{\sum f_{i}d_{i}}{\sum f_{i}}=\frac{12}{30}=0.4\) \(\bar{x}=\bar{d}+a=0.4+75.5=75.9\)

 

Question 5:

The oranges were packed and were sold by the fruit seller in a market. The numbers of oranges are arranged in different boxes by the fruit seller. Through the number of oranges the number of oranges are distributed in following manner.

No.of oranges	50-52	53-55	56-58	89-61	62-64
No. of boxes	15	110	135	115	25

In the distributed pattern the number of oranges. Find the mean. Choose the method to find the mean:

Solution:

Class interval	fi	xi	di=x-a	fi di
50-52	110	54	-6	90
53-55	15	51	-3	-330
56-58	135	60	0	0
59-61	25	57	3	345
62-64	115	63	6	150
	Sum fi= 400			Sum fi di=75

Mean can be calculated as follows:

\(\bar{d}=\frac{\Sigma f_{i}d_{i}}{\Sigma f_{i}}=\frac{75}{400}=0.1875\) \(\bar{x}=\bar{d}+a=0.1875+57=57.1875\)

In this case, there are wide variations in \(f_{i}\) and hence assumed mean method is used.

 

Question 6:

The following distribution shows the food expenditure of the given households from the locality:

Daily food expenditure (in Rs)	50-100	100-150	150-200	200-250	250-300
Number of given households	4	5	12	2	2

Find the mean daily expenditure on food by a suitable method.

Solution:

Class Interval	fi	xi	di= xi – a	ui = di/h	fiui
50-100	4	125	-100	-2	-8
100-150	5	175	-50	-1	-5
150-200	12	225	0	0	0
200-250	2	275	50	1	2
250-300	2	325	100	2	4
	Sum fi = 25				Sum fiui = -7

Mean can be calculated as follows:

\(\bar{x}=a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}\times h\) \(=225+\frac{-7}{25}\times 50 =225\)

 

Question 7:

To find concentration in sulphur dioxide (in parts per million, i.e. ppm),The parts of data are collected by different areas from cities are given in distributed table:

Concentration in SO2 (in ppm)	Frequency of F
0.00-0.04	4
0.04-0.08	9
0.08-0.12	9
0.12-0.16	2
0.16-0.20	4
0.20-0.24	2

Give the mean of the sulphur dioxide collected from the given cities.

Solution:

Class Interval	fi	xi	fixi
0.00-0.04	9	0.02	0.08
0.04-0.08	4	0.06	0.54
0.08-0.12	9	0.10	0.90
0.12-0.16	2	0.14	0.28
0.16-0.20	2	0.18	0.72
0.20-0.24	4	0.22	0.44
	Sum fi=30		Sum fixi=2.96

Mean can be calculated as follows:

\(\bar{x}=\frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}=\frac{2.96}{30}=0.099 ppm\)

 

Question 8:

The absentee record of 40 students is handled by the teacher for one complete term. Find the mean of absent days of the given students:

Number of following days	0-6	6-10	10-14	14-20	20-28	28-38	38-40
Number of following students	10	11	4	7	3	4	1

Solution:

Class Interval	fi	xi	fixi
0-6	11	3	33
6-10	10	8	80
10-14	4	12	84
14-20	7	17	68
20-28	3	24	96
28-38	4	33	99
38-40	1	39	39
	Sum fi = 40		Sum fixi= 499

Mean can be calculated as follows:

\(\bar{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}\)= \(\frac{499}{40}\)=12.4

 

Question 9:

The data of the literacy digit from the 35 cities is given in the following table. Find the mean of the 35 literacy cities.

Literacy rate from 35cities (in %)	40-45	50-55	60-65	70-75	80-85
Cities in values	10	3	8	11	3

Solution:

Class Interval	fi	xi	di=xi-a	ui=di/h	fiui
40-45	10	50	-20	-2	-6
50-55	3	60	-10	-1	-10
60-65	8	70	0	0	0
70-75	11	80	10	1	8
80-85	3	90	20	2	6
	Sum fi=35				Sum fiui= -2

Mean can be calculated as follows:

\(\bar{x}=a+\frac{\Sigma f_{i}u_{i}}{\Sigma f_{i}}\times h\) \(= 70+\frac{-2}{35}\times 10=69.42\)

 

EXERCISE – 2

 

Question 1:

The age of patients admitted in a hospitals are given in the following table for a complete 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

 

Calculate the mode and mean of the data above. Compare and interpret the two measures of central tendency.

Solution:

Modal class=35-45, length = 35, height = 10, \( a_{1} = 23, a_{0} = 21\, and\, a_{3} = 14\)

 

Therefore, Mode = \(l +\frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}}\times h\)

= \(35+\left ( \frac{23-21}{2\times 23-21-14} \right )\times 10\)

= \(35+\frac{2}{11}\times 10=36.8\)

Calculating Mean

CLASS INTERVAL	ai	zi	aizi
5-15	6	10	60
15-25	11	20	220
25-35	21	30	630
35-45	21	30	920
45-55	14	50	700
55-65	5	60	300
	Sum ai = 80		Sum aizi= 2830

 

\(\bar{p}=\frac{\sum a_{i}x_{i}}{\sum a_{i}}=\frac{2830}{80}=35.37\)

 

\(\ therefore\)The mode of data displays that the maximum number of patients is in the age group of 26.8, whereasthe average age of all the patients is 35.37.

 

Question 2:

The data given below provides the information for the observed lifetime (in hours) of 225 electrical components:

Lifetime ( in hours)	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100	100 – 120
Frequency	10	35	52	61	38	29

Find the modal lifetimes of the components.

 

Solution:

Modal class=60-80, l = 60,\( a_{1} = 61, a_{0} = 52, a_{2} = 38\) and h = 20

\(Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h\)

 

\(= 60 + \left ( \frac{61 – 52}{2 \times 61 – 52 – 38 } \right ) \times 20\)

 

\(= 60 + \frac{9}{32} \times 20 = 65.62\)

 

Question 3:

The total monthly household expenses of 200 families of a town are given in the following table. Calculate the modal monthly expenses of the families. Also, calculate the mean monthly expenses.

Expenses	Number of families
1000 – 1500	24
1500 – 2000	40
2000 – 2500	33
2500 – 3000	28
3000 – 3500	30
3500 – 4000	22
4000 – 4500	16
4500 – 5000	7

 

Solution:

Modal class=1500-2000, l = 1500,\( a_{1} = 40, a_{0} = 24, a_{2} = 33\) and h = 500

\(Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h\)

 

\(= 1500 + \left ( \frac{40 – 24}{2 \times 40 – 24 – 33} \right ) \times 500\)

 

\(= 1500 + \left ( \frac{16}{23} \right ) \times 500 = 1847.82\)

Calculation of mean:

Class Interval	ai	zi	di =zi – b	ui= di/h	aiui
1000 – 1500	24	1250	-1500	-3	-72
1500 – 2000	40	1750	-1000	-2	-80
2000 – 2500	33	2250	-500	-1	-33
2500 – 3000	28	2750	0	0	0
3000 – 3500	30	3250	500	1	30
3500 – 4000	22	3750	1000	2	44
4000 – 4500	16	4250	1500	3	48
4500 – 5000	7	4750	2000	4	28
	ai = 200				aiui = -35

 

\(\bar{p}=\frac{\sum a_{i}x_{i}}{\sum a{i}}\times h\) \(= 2750 – \frac{35}{200} \times 500\)

 

= 2750 – 87.5 = 2662.50

 

Question 4:

The state – wise teacher- student ratio in a college in India are given in the following table. Calculate the mean and mode of the given data. Interpret the two measures.

 

Number of students per teacher	Number of states/UT
15 – 20	3
20 – 25	8
25 – 30	9
30 – 35	10
35 – 40	3
40 – 45	0
45 – 50	0
50 – 55	2

 

Solution:

Modal class=30-35, l = 30, \(a_{1} = 10, a_{0} = 9, a_{2} = 3\) and h = 5

 

\(Mode = l +\left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h\)

 

\(= 30 – \left ( \frac{20 – 9}{2 \times 10 – 9 – 3 } \right ) \times 5\)

 

\(= 30 – \frac{1}{8} \times 5 = 30.625\)

 

Calculating mean:

Number of students per teacher	ai	zi	di= zi –b	ui = ui/h	aiui
15 – 20	3	17.5	-15	-3	-9
20 – 25	8	22.5	-10	-2	-16
25 – 30	9	27.5	-5	-1	-9
30 – 35	10	32.5	0	0	0
35 – 40	3	37.5	5	1	3
40 – 45	0	42.5	10	2	0
45 – 50	0	47.5	15	3	0
50 – 55	2	52.5	20	4	8
	Sum ai = 35				sum aiui= -23

\(\bar{p}=\frac{\sum a_{i}x_{i}}{\sum a_{i}}\times h\)

 

\(= 30.5 – \frac{23}{35}\times 5\)

 

\(= 30.5 – \frac{23}{7} = 29.22\)

 

Therefore, from the mode we know that the maximum number of states has 30 – 35 students per teacher, and from the mean we know that the average ratio of students per teacher is 29.22

 

Question 5:

Provided table shows the total runs scored by some of the top batsman of the world in one-day cricket matches.

Runs Scored	Number of Batsman
3000 – 4000	4
4000 – 5000	18
5000 – 6000	9
6000 – 7000	7
7000 – 8000	6
8000 – 9000	3
9000 – 10000	1
10000 – 11000	1

Calculate the mode of the given information.

Solution:

Modal class=4000-5000, l = 4000, \(a_{1} = 18, a_{0} = 4, a_{2} = 9\) and h = 1000

\(Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h\) \(= 4000 + \left ( \frac{18 – 4}{2 \times 18 – 4 -9} \right )\times 1000\) \(= 4000 + \left ( \frac{14}{23} \right )\times 1000 = 4608.70\)

 

Question 6:

A complete observation of the number of cars on the roads for 100 periods of 3 minutes are summarized in the following table. Calculate 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	8

 

Solution:

Modal class=40 – 50, l = 40, \(a_{1} = 20, a_{0} = 12, a_{2} = 11\) and h = 10

\(Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h\)

 

\(= 40 + \left ( \frac{20 – 12}{2 \times 20 – 12 – 11} \right )\times 10\)

 

\(= 40 + \left ( \frac{8}{17} \right )\times 10 = 44.70\)

 

 

Exercise 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 \(\frac{n}{2}=34\)

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

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

=125+12=137

 

Mode = Modal class=125-145, \(f_{1} = 20, f_{0} = 13, f_{2} = 14\) & h = 20

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

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

=\(125+\frac{7}{13}X20\)

=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

 

\(\bar{x}=a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}X h\) \(=135+\frac{7}{68}X20\)

=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,\(\frac{n}{2}\) = 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+\(\left ( \frac{\frac{n}{2}-cf}{f} \right )Xh\)

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

Or,\(\frac{25-x}{2}\)

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 \(\frac{n}{2}\) = 50

Median class=35-45

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

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

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

=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 \(\frac{n}{2}\) = 20

Median class=144.5-153.5

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

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

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

=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 &\(\frac{n}{2}\) = 200

Median class=3000 – 3500

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

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

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

=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 &\(\frac{n}{2}\) = 50

Median class=7-10

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

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

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

=8.05

Calculate the Mode:

Modal class=7-10,

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

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

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

=\(7+\frac{10}{34}X3\)

=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

 

\(\bar{x}=a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}X h\) \(=\frac{825}{100}\)

=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 \(\frac{n}{2}\)= 15

median class=55-60

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

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

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

=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, \(x_{i}\) = 18(given)

Hence substituting for f_ix_i,

 

18(44+f) = 752 + 20f

Or, 792 + 18f = 752 + 20f

Or, 2f = 40

\( \rightarrow f=20\)