Andhra Pradesh SSC Class 10 Solutions For Maths – Statistics

Andhra Pradesh SSC Class 10 Solutions For Maths – Statistics (English Medium)

These Solutions are part of AP SSC Class 10 Solutions for Maths. In this article, we have provided Andhra Pradesh SSC Class 10 Solutions For Maths Chapter 14 Statistics

Exercise 14.1:

Question 1:
A 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.
Solution :
Let us find the class marks x_i of each class by taking the average of upper class limit and lower class limit and put them in a table.

Thus, the mean number of plants per house is 8.1 plants
We have used direct method because numerical values of x_i and f_i are small.

AP SSC 10th Class Textbook Solutions

Question 2:
Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution :
Let us find the class marks x_i of each class by taking the average of upper class limit and lower class limit and put them in a table.
Let us assume a=325 and h=50

Class interval	No. of students(fi)	Class marks (xi)	di= xi-325		fiui
200-250	12	225	-100	-2	-24
250-300	14	275	-50	-1	-14
300-350	8	325	0	0	0
350-400	6	375	50	1	6
400-450	10	425	100	2	20
Total	∑fi=50				-12

Using step-deviation method

Thus, the mean daily wages is Rs. 313.
*Note: Answer given in the textbook is incorrect.

Question 3:
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is 18. Find the missing frequency ƒ.

Solution :
Let us find the class marks x_i of each class by taking the average of upper class limit and lower class limit and put them in a table. Let us assume a=18 and h=2

Class interval	No. of students(fi)	Class marks (xi)	di=xi-a		fiui
11-13	7	12	-6	-3	-21
13-15	6	14	-4	-2	-12
15-17	9	16	-2	-1	-9
17-19	13	18	0	0	0
19-21	f	20	2	1	f
21-23	5	22	4	2	10
23-25	4	24	6	3	12
Total	∑fi=44+f				∑fiui=-20+f

The missing frequency is 20.

Question 4:
Thirty women were examined in a hospital by a doctor and their of heart beats per minute were recorded and summarised as shown. Find the mean heart beats per minute for these women, choosing a suitable method.

Solution :
Let us find the class marks x_i of each class by taking the average of upper class limit and lower class limit and put them in a table.
Let us assume a=75.5 and h=3

Class interval	No. of students(fi)	Class marks (xi)	di=xi-a		fiui
65-68	2	66.5	-9	-3	-6
68-71	4	69.5	-6	-2	-8
71-74	3	72.5	-3	-1	-3
74-77	8	75.5	0	0	0
77-80	7	78.5	3	1	7
80-83	4	81.5	6	2	8
83-86	2	84.5	9	3	6
Total	∑fi=30				∑fiui=4

The mean heart beats per minute is 75.9

Question 5:
In a retail market, fruit vendors were selling oranges kept in packing baskets. These baskets contained varying number of oranges. The following was the distribution of oranges.

Find the mean number of oranges kept in each basket. Which method of finding the mean did you choose?
Solution :
Let us find the class marks x_i of each class by taking the average of upper class limit and lower class limit and put them in a table.

Class interval	No. of students(fi)	Class marks (xi)	fixi
10-14	15	12	180
15-19	110	17	1870
20-24	135	22	2970
25-29	115	27	3105
30-34	25	32	800
Total	∑fi=400		∑fixi=8925

The mean number of oranges kept in basket is 22.31.We have used direct method.

Question 6:
The 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.
Solution :
Let us find the class marks x_i of each class by taking the average of upper class limit and lower class limit and put them in a table.

Class interval	No. of students(fi)	Class marks (xi)	fixi
100-150	4	125	500
150-200	5	175	875
200-250	12	225	2700
250-300	2	275	550
300-350	2	325	650
Total	∑fi=25		∑fixi=5275

The mean daily expenditure on food of 25 household in a locality is Rs.211

Question 7:
To find out the concentration of S0₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO₂ in the air.
Solution :
Let us find the class marks x_i of each class by taking the average of upper class limit and lower class limit and put them in a table.

Class interval	No. of students(fi)	Class marks (xi)	fixi
0.00-0.04	4	0.02	0.08
0.04-0.08	9	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	4	0.18	0.72
0.20-0.24	2	0.22	0.44
Total	∑fi=30		∑fixi=2.96

Question 8:
A class teacher has the following attendance record of 40 students of a class for the whole term. Find the mean number of days a student was present out of 56 days in the term.

Solution :
Let us find the class marks x_i of each class by taking the average of upper class limit and lower class limit and put them in a table.
Let us assume a=45.5 and h=3

Class interval	No. of students(fi)	Class marks (xi)	di=xi-a		fiui
35-38	1	36.5	-9	-3	-3
38-41	3	39.5	-6	-2	-6
41-44	4	42.5	-3	-1	-4
44-47	4	45.5	0	0	0
47-50	7	48.5	3	1	7
50-53	10	51.5	6	2	20
53-56	11	54.5	9	3	33
Total	∑fi=40				∑fiui=47

Thus, the mean number of days a student was present out of 56 days in the term is 49.

Question 9:
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution :
Let us find the class marks x_i of each class by taking the average of upper class limit and lower class limit and put them in a table.

Class interval	No. of students(fi)	Class marks (xi)	fixi
45-55	3	50	150
55-65	10	60	600
65-75	11	70	770
75-85	8	80	640
85-95	3	90	270
Total	∑fi=35		∑fixi=2430

The mean literacy rate is 69.43%.

Exercise 14.2:

Question 1:
The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution :
Let us find the class marks x_i of each class, and put them in a table.

Class interval	No. of patient(fi)	Class marks (xi)	fixi
5-15	6	10	60
15-25	11	20	220
25-35	21	30	630
35-45	23	40	920
45-55	14	50	700
55-65	5	60	300
Total	∑fi=80		∑fixi=2830

Question 2:
The following data gives the information on the observed life times (in hours) of 225 electrical components :

Determine the modal lifetimes of the components.
Solution :

Question 3:
The following data gives the distribution of total monthly household expenditure of 200 families of Gummadidala village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

Solution :

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

Solution :

Question 5:
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.
Solution :

Question 6:
A student noted the number of cars passing through a spot on a road for 100 periods, each of 3 minutes, and summarised this in the table given below.

Find the mode of the data.
Solution :

Exercise 14.3:

Question 1:
The 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.

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

Mean:

Class interval	No. of consumer(fi)	Class marks (xi)	fixi
65-85	4	75	300
85-105	5	95	475
105-125	13	115	1495
125-145	20	135	2700
145-165	14	155	2170
165-185	8	175	1400
185-205	4	195	780
Total	∑fi=68		∑fixi=9320

The three measures are approximately the same in this case.

Question 2:
If the median of 60 observations, given below is 28.5, find the values of x and y.

Solution :

Question 3:
A life insurance agent found the following data about distribution of ages of 100 policy holders. Calculate the median age. [Policies are given only to persons having age 18 years onwards but less than 60 years.]

Solution :
To calculate the median age, we need to find the class intervals and frequency.

Class interval	frequency	Cumulative frequency
18-20	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

To round 13.73 to nearest tenth means to round the numbers so you only have one digit in the fractional part.

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

Find the median length of the leaves. (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5,…. ,171.5 – 180.5.)
Solution :
Observe the given table.
The upper limit of the previous class is not equal to the lower limit of the consequent class.
This table is called inclusive type of frequency distribution table.
And the data
The data needs to be converted to continuous classes for finding the median.
So change the classes from 118-126,127-135,136-144,145-153,154-162,163-171 and 172-180 to 117.5-126.5, 126.5-135.5, 135.5
144.5, 144.5-153.5, 153.5-162.5, 162.5-171.5 and 171.5-180.5 respectively.
And hence the class size changes to 9.
Now let us write the frequency distribution table.

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

Question 5:
The following table gives the distribution of the life-time of 400 neon lamps Find the median life time of a lamp.

Solution :
Consider the following frequency distribution table:

Class interval	frequency	Cumulative frequency
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

Question 6:
100 sumames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows

Determine the median number of letters in the sumames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Solution :
Median
Consider the following frequency distribution table:

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

Mean:

Class interval	No. of surnamesr(fi)	Class marks (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	58
16-19	4	17.5	70
Total	∑fi=100		∑fixi=832

Question 7:
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Solution :
Median
Consider the following frequency distribution table:

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

Exercise 14.4:

Question 1:
The following distribution gives the daily income of 50 workers of a factory.

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Solution :
The frequency distribution table of less than type:

Daily income(in rupees)	No. of workers
Less than 300	12
Less than 350	26
Less than 400	34
Less than 450	40
Less than 500	50

We first draw the coordinate axes, with upper limits of the income along the horizontal axis, and the cumulative frequency along the vertical axes. Then we plot the points (300,12), (350,26), (400,34), (450,40) and (500,50). We join these points with a smooth curve to get the less than type ogive, as shown in the figure below.

Note: The answer given in the text book is incorrect.

Question 2:
During the medical check-up of 3 5 students of a class, their weights were recorded as follows :

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verity the result by using the formula.
Solution :
We first draw the coordinate axes, with upper limits of the weight along the horizontal axis, and the cumulative frequency along the vertical axes. Then we plot the points (38,0), (40,3), (42,5), (44,9), (46,14), (48,28), (50,32) and (52,35). We join these points with a smooth curve to get the less than type ogive, as shown in the figure below.

Here n=35
So, n/2=17.5
Now draw line y=17.5 parallel to x-axis.
Let the line y=17.5 meets the ogive at the point A.
Mark the point A.
From draw a line parallel to y-axis.
Let this line meets the x-axis at x=46.5
Therefore the median is 46.5
Now let us verify the result by using the formula.
It was observed that the difference between two consecutive upper class limits is 2. Let us construct the frequency distribution table.

weight	frequency	Cumulative frequency
Less than 38	0	0
38-40	3-0=3	3
40-42	5-3=2	5
42-44	4	9
44-46	5	14
46-48	14	28
48-50	4	32
50-52	3	35
Total(n)	35

Question 3:
The following table gives production yield per hectare of wheat of 100 farms of a village.

Change the distribution to a more than type distribution, and draw its ogive.
Solution :
The cumulative frequency distribution of more than type is:

Production yield	No. of farmers
More than or equal to 50	100
More than or equal to 55	100-2=98
More than or equal to 60	98-8=90
More than or equal to 65	90-12=78
More than or equal to 70	78-24=54
More than or equal to 75	54-38=16

We first draw the coordinate axes, with lower limits of the production yield along the horizontal axis, and the cumulative frequency along the vertical axes. Then we plot the points (50,100), (55,98), (60,90), (65,78), (70,54) and (75,16). We join these points with a smooth curve to get the more than type ogive, as shown in the figure below.

AP Board Solutions

Hope given Andhra Pradesh SSC Class 10 Solutions For Maths chapter 14 Statistics are helpful to complete your math homework.

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