NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability
NCERT Exemplar class 10 maths solutions chapter 13 is an extension to the learnings of statistics and probability done in class 9. The chapter on Statistics and Probability is of great importance for the examinations as well as future career prospects for a student. All the data analytics and artificial intelligencerelated verticals require a sound knowledge of Statistics and Probability. The NCERT exemplar class 10 maths chapter 13 solutions are curated by our highly experienced content development team which enables the students to study and practice NCERT class 10 Maths. effectively. These NCERT exemplar class 10 maths chapter 13 solutions explore the basic and intermediate level concepts of statistics and probability and follow the CBSE 10th maths syllabus via NCERT exemplar class 10 maths solutions chapter 13.
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NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability: Exercise13.1
Question:1
In the formula
for finding the mean of grouped data diâ€™s are deviations from
(A) lower limits of the classes
(B) upper limits of the classes
(C) mid points of the classes
(D) frequencies of the class marks
Answer:
Answer. [C]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate mean. First of all, add up all the observations and then divide
by the total number of observations.
We know that d_{i} = x_{i} â€“ a
where x_{i} is data and a is mean
So, di are the derivative from midpoint of the classes.
Question:2
While computing mean of grouped data, we assume that the frequencies are
(A) evenly distributed over all the classes
(B) centred at the class marks of the classes
(C) centred at the upper limits of the classes
(D) centred at the lower limits of the classes
Answer:
Answer. [B]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate mean. First of all, add up all the observations and then divide by the total number of observations.
Hence while computing mean of grouped data, we assume that the frequencies are centered at the class marks of the classes
Question:3
If x_{i}â€™s are the mid points of the class intervals of grouped data, fiâ€™s are the corresponding frequencies and is the mean, then is equal to
(A) 0 (B) â€“1 (C) 1 (D) 2
Answer:
Answer. [A]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate mean. First of all, add up all the observations and then divide by the total number of observations.
That is mean
By cross multiplication we get
(from equation (1))
= 0
Question:4
In the formula for finding the mean of grouped frequency distribution, u_{i} =
(A) (B) h(xi â€“ a) (C) (D)
Answer:
Answer. [C]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate mean. First of all, add up all the observations and then divide by the total number of observations.
Also we know that d_{i} = x_{i} â€“ a and
put d_{i} = x_{i} â€“ a
Hence option C is correct
Question:5
The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its
(A) mean (B) median (C) mode (D) all the three above
Answer:
Answer. [B]
Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
If we make graph of less than type and of more than type grouped data and find the intersection point then the value at abscissa is the median of the grouped data.
Hence option (B) is correct.
Question:6
For the following distribution :
the sum of lower limits of the median class and modal class is
(A) 15 (B) 25 (C) 30 (D) 35
Answer:
Answer. [B]
Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
Class  Frequency  Cumulative Frequency (C.F) 
05  10  10 
510  15  (10+15=25) 
1015  12  (25+12=37) 
1520  20  (37+20=57) 
2025  9  (57+9=66) 
N=66 
which ies in the class 1015.
Hence the median class is 10 â€“ 15
The class with maximum frequency is modal class which is 15 â€“ 20
The lower limit of median class = 10
The lower limit of modal class = 15
Sum = 10 + 15 = 25
Question:7
Consider the following frequency distribution:
The upper limit of the median class is
(A) 17 (B) 17.5 (C) 18 (D) 18.5
Answer:
Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
Class in not continuous. So we have to make 1t continuous first.
Class  Frequency  Cumulative Frequency 
05.5  13  13 
5.511.5  10  (13+10=23) 
11.517.5  15  (23+15=38) 
17.523.5  8  (38+8=46) 
23.529.5  11  (46+11=57) 
N = 57 
Here the median class is 15.5 â€“ 17.5
upper limit of median class is 17.5
Question:8
For the following distribution :
the modal class is
(A) 1020 (B) 2030 (C) 3040 (D) 5060
Answer:
Answer. [C]
Solution.
Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to
summarize categorical variables.
Marks  Frequency  Cumulative Frequency 
010  3  3 
1020  123=9  12 
2030  2712=15  27 
3040  5727=30  57 
4050  7557=18  75 
5060  8075=5  80 
=80 
The class with highest frequency is 3040
Hence 30 â€“ 40 is the modal class.
Question:9
The difference of the upper limit of the median class and the lower limit of the modal class is
(A) 0 (B) 19 (C) 20 (D) 38
Answer:
Answer. [C]
Solution. Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
Class  Frequency  Cumulative Frequency 
6585  4  4 
85105  5  (4+5=9) 
105125  13  (9+13=22) 
125145  20  (22+20=42) 
145165  14  (42+14=56) 
165185  7  (56+7=63) 
185205  4  (63+4=67) 
N = 67 
Median class = 125 â€“ 145
upper limit of median = 145
The class with maximum frequency is modal class which is 125 â€“ 145
lower limit of modal class = 125
Difference of the upper limit of median and lower limit of modal = 145 â€“ 125 = 20
Question:10
The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below :
The number of athletes who completed the race in less than 14.6 seconds is :
(A) 11 (B) 71 (C) 82 (D) 130
Answer:
Answer. [C]
Solution. Frequency: The number of times a event occurs is a specific period is called frequency.
The number of athletes who are below 14.6 = frequency of class (13.814) + frequency of class (14 14.2) +
frequency of class (14.214.4) + frequency of class (14.414.6)
= 2 + 4 + 5 + 71 = 82
Hence the frequency of race completed in less than 14.6 = 82
Question:11
Consider the following distribution :
the frequency of the class 3040 is
(A) 3 (B) 4 (C) 48 (D) 51
Answer:
Answer. [A]
Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use
frequency distribution to summarize categorical variables.
Marks obtained  Cumulative Frequency  Frequency 
010  63  5 
1020  58  3 
2030  55  4 
3040  51  3 
4050  48  6 
5060  42  42 
So the frequency of class 30 â€“ 40 is 3.
Question:12
If an event cannot occur, then its probability is
(A) 1 (B) (C) (D) 0
Answer:
Answer. [D]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Here number of favorable cases is 0.
Probability =
Probability =
Question:13
Which of the following cannot be the probability of an event?
(A) (B) 0.1 (C) 3% (D)
Answer:
Answer. [D]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
(A) 1/3
Here 0 < 1/3 < 1
Hence it can be the probability of an event.
(B) 0.1
Here 0 < 0.1 < 1
Hence it can be the probability of an event.
(C) 3% = 3/100 = 0.03
Here 0 < 0.03 < 1
Hence it can be the probability of an event.
(D)17/16
Here
Hence is not a probability of event
Hence option (D) is correct answer.
Question:14
An event is very unlikely to happen. Its probability is closest to
(A) 0.0001 (B) 0.001 (C) 0.01 (D) 0.1
Answer:
Answer. [A]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The descending order of option (A), (B), (C), (D) is
0.1 > 0.01 > 0.001 > 0.0001 that is (D) > (C) > (B) > (A)
We can also say that it is the order of happening of an event.
Here 0.0001 it is the smallest one.
Hence 0.0001 is very unlikely to happen
Question:15
If the probability of an event is p, the probability of its complementary event will be
(A) p â€“ 1 (B) p (C) 1 â€“ p (D)
Answer:
Answer. [C]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The probability of an event = p
Let the probability of its complementary event = q
We know that total probability is equal to 1.
Hence, p + q = 1
q = 1 â€“ p
Question:16
The probability expressed as a percentage of a particular occurrence can never be
(A) less than 100
(B) less than 0
(C) greater than 1
(D) anything but a whole number
Answer:
Answer. [B]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The probability expressed as presentage of an event A is btewwen 0 to 100.
Hence we can say that probability can never be less than 0.
Hence option (B) is correct.
Question:17
If P(A) denotes the probability of an event A, then
(A) P(A) < 0 (B) P(A) > 1 (C) 0 â‰¤ P(A) â‰¤ 1 (D) â€“1 â‰¤ P(A) â‰¤ 1
Answer:
Answer. [C]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
(A) P(A) < 0
It is not represent the probability of event A because probability of an event can never be less than 0.
(B) P(A) > 1
It is not represent the probability of event A because probability of an event can never be greater than 1.
(C) 0 â‰¤ P(A) â‰¤ 1
It represent probability of event A because probability of an event is always lies from 0 to 1.
(D) â€“1 â‰¤ P(A) â‰¤ 1
It is not represent the probability of event A because probability of an event can never be equal to 1.
Hence option (C) is correct .
Question:18
A card is selected from a deck of 52 cards. The probability of its being a red face card is
(A) (B) (C) (D)
Answer:
Answer. [A]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total number of cases = 52
Red face cards = 6
Favorable cases = 6
Let event A is to select a card from 52 card.
Probability that it is a red card is p(A)
Question:19
The probability that a non leap year selected at random will contain 53 Sundays is
(A) (B) (C) (D)
Answer:
Answer. [A]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
In 365 days there are 52 weeks and 1 day.
If it contain 53 sunday then the 1 day of the year must be sunday.
But there are total 7 days.
Hence total number of favorable cases = 1
Hence probability of 53 sunday =
Question:20
When a die is thrown, the probability of getting an odd number less than 3 is
(A) (B) (C) (D) 0
Answer:
Answer. [A]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total no. of cases = 6
odd number less than 3 = 1
Number of favorable cases = 1
Probability =
Question:21
A card is drawn from a deck of 52 cards. The event E is that card is not an ace of hearts. The number of outcomes favourable to E is
(A) 4 (B) 13 (C) 48 (D) 51
Answer:
Answer. [D]
Solution. Total number of cards = 52
Ace of hearts = 1
The card is not an ace of hearts = 52 â€“ 1 = 51
The number of outcomes favourabe to E = 51
Question:22
The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is
(A) 7 (B) 14 (C) 21 (D) 28
Answer:
Answer. [B]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Let event A is to get a bad egg.
So, p (A) = 0.035 (given)
P(A) =
0.035 =
Number of favourable cases =
Question:23
A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets has she bought?
(A) 40 (B) 240 (C) 480 (D) 750
Answer:
Answer. [C]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases = 6000
Probability of getting first prize (p(A)) = 0.08
p(A)
0.08 Ã— 6000 = Number of tickets the bought
= Number of tickets the bought
Number of tickets the bought = 480.
Question:24
Answer:
Answer. [A]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total tickets = 40
Number of tickets multiple of 5 = 5, 10, 15, 20, 25, 30, 35, 40
Total favourable cases = 8
Let A be the event of getting a ticket with number multiple of 5.
p(A) =
Question:25
Someone is asked to take a number from 1 to 100. The probability that it is a prime is
(A) (B) (C) (D)
Answer:
Answer. [C]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total number of cases = 100
prime number from 1 to 100 = 2, 3, 5, 7, 9, 11, 13, 17, 19,
23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 83, 89, 97
Total prime numbers from 1 to 100 = 25
Probability of getting prime number =
Question:26
Answer:
Answer. [B]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a
random event. The value is expressed from zero to one.
Total students = 23
Students in A, B, C = 4 + 8 + 5 = 17
Students in C, D = 23 â€“ 17 = 6
Number of favourable cases = 6
Let A be the event that the student is not from A, B, C
NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability: Exercise13.2
Question:1
Answer:
Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
Ungrouped data: The data which is not grouped is called ungrouped data.
The median is the middle number in the grouped data but when data is ungrouped the median is also changed.
Hence the median is not same of grouped and ungrouped data
Question:2
Answer:
Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
Mean : It is the average of the given numbers. It is easy
to calculate mean. First of all add up all the numbers then divide by how many numbers are there.
This last statement is not correct because a can be any point in the grouped data it is not necessary that a must be midpoint.
Hence the statement is false.
Question:3
Answer:
Answer. [False]
Solution. Mean : It is the average of the given numbers. It is easy to calculate mean. First of all add up all the numbers then divide by how many numbers are there.
Grouped data are data formed by aggregating individual observations of a variable into groups.
The mean, mode and median of grouped data can be the same it will depend on what type of data is given.
Hence the statement is false.
Question:4
Will the median class and modal class of grouped data always be different? Justify your answer.
Answer:
Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
The median is always the middle number and the modal class is the class with highest frequency it can be happen that the median class is of highest frequency.
So the given statement is false median class and mode class can be same.
Question:5
Answer:
Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total children = 3
Cases â€“ GGG, GGB, GBG, BGG, BBB, BBG, BGB, GBB were G is girl and B is boy.
Probability =
Probability of 0 girl =
Probability of 1 girl =
Probability of 2 girl =
Probability of 3 girl =
Here they are not equal to
Question:6
Answer:
Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Here 3 contain 50% of the region and 1, 2, contain 25%, 15% of the region.
All probabilities are not equal. So the given statement is false.
Question:7
Answer:
Answer. [Peehu]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
As apoorv throws two dice total cases = 36
Product is 36 when he get = (6, 6)
Number of favourable cases = 1
Probability =
Probability that Apoorv get 36 =
Peehu throws are die total cases = 6
Square of 6 is 36
Hence case = 1
Probability that Peehu get 36 =
Hence Peehu has better cases to get 36.
Question:8
Answer:
Answer. [True]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases when we toss a coin = 2(H, T)
Probability =
Probability of head =
Probability of tail =
Hence the probability of each outcome is .
Question:9
Answer:
Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Here total cases = 6
Number of favourable cases in getting 1 = 1
Probability =
Probability of getting
Number of favourable cases 'not 1' = 5 (2, 3, 4, 5, 6)
Probability of not 1 =
Hence they are not equal to
Question:10
Answer:
Answer. [ ]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases in tossing three coins = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
Number of case with no head = TTT
Probability =
Probability of no head =
The conclusion that probability of no head is is wrong because as we calculate it above, it comes out . Hence the probability of no head is
Question:11
Answer:
Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The probability of getting a head is 1, means that we never get tail. But this is not true because we have both head and tail in a coin. Hence probability of getting head is 1 is false.
Question:12
Answer:
Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
No, because when we toss a coin we can get either tail or head and the probability of each is .
So, it is not necessary that she gets tail at fourth toss. She can get head also.
Question:13
Answer:
Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
No, because we get head or tail after tossing a coin that is the probability of both outcomes is .
Hence tail is not have higher chance than head.
Both are have equal chance.
Answer:
Answer. [True]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total slips = 100
Slips with even number = 50
Probability =
Probability of slip with even number =
Slips with odd number = 50
Probability of slip with odd number =
Hence the probability of each is .
NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability: Exercise13.3
Question:1
Find the mean of the distribution :
Answer:
Answer. [5.5]
Solution. Here we calculate mean by following the given steps:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
Class
Marks (x_{i})
Frequency(f_{i})
f_{ix}i
13
2
9
18
35
4
22
88
57
6
27
162
710
8.5
17
144.5
Question:2
Calculate the mean of the scores of 20 students in a mathematics test :
Answer:
Answer. [35]
Solution. Here we calculate mean by following the given steps:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
4 Now divide sum of (fx) by sum of (f)
Marks  x_{i}  No.of students fi  f_{ix}i 
1020  15  2  30 
2030  25  4  100 
3040  35  7  245 
4050  45  6  270 
5060  55  1  55 
Question:3
Calculate the mean of the following data
Answer:
Answer. [12.93]
Solution. Here we calculate mean by following the given steps:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
Class  x_{i}  fi  f_{i x}i 
47  5.5  5  275 
811  9.5  4  38 
1215  13.5  9  121.5 
1619  17.5  10  175 
Question:4
Find the mean number of pages written per day.
Answer:
Answer. [26]
Solution. Here we calculate mean by following the given steps:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
No.of pages written per day
_{no.of days(fi)}
(xi)
f_{ix}i
1618
1
17
17
1921
3
20
60
2224
4
23
92
2527
9
26
234
2830
13
29
377
Question:5
The daily income of a sample of 50 employees are tabulated as follows :
Find the mean daily income of employees.
Answer:
Answer. [356.5]
Solution. Here we calculate mean by following the given steps:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
Income (in Rs )
xi
No.of employees
f_{ix}i
1200
100.5
14
1407
201400
300.5
15
4507.5
401600
500.5
14
7007
601800
700.5
7
4903.5
Question:6
Determine the mean number of seats occupied over the flights.
Answer:
Answer. [109]
Solution. Here we calculate mean by following the given steps:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
Number of seats  Frequency fi  xi  f_{ix}i 
100104  15  102  1530 
104108  20  106  2120 
108112  32  110  3520 
112116  18  114  2052 
116120  15  118  177065268 
number of seats = 109
Question:7
The weights (in kg) of 50 wrestlers are recorded in the following table :
Find the mean weight of the wrestlers.
Answer:
Answer. [123.4]
Solution. Here we calculate mean by following the given steps:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
Weight
fi
xi
f_{ix}i
100110
4
105
420
110120
14
115
1610
120130
21
125
2625
130140
8
135
1080
140150
3
145
435
Question:8
Answer:
Answer. [14.48]
Solution. Here we calculate mean by following
1.Find the mid point of each interval.
2.Multiply the frequency of each interval by its mid point.
3.Get the sum of all the frequencies (f) and sum of all the (fx)
4.Now divide sum of (fx) by sum of (f)
MIleage (km/I)  No.of cars (fi)  xi  f_{ix}i 
1012  7  11  77 
1214  12  13  156 
1416  18  15  270 
1618  13  17  221 

Question:9
The following is the distribution of weights (in kg) of 40 persons :
Construct a cumulative frequency distribution (of the less than type) table for the data above.
Answer:
Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
CI.  f  cf 
4045  4  4 
4550  4  8 
5055  13  21 
5560  5  26 
6065  6  32 
6570  5  37 
7075  2  39 
7580  1  40 
Question:10
Construct a frequency distribution table for the data above.
Answer:
Solution. Frequency distribution: It tells how frequencies are distributed over values in a
frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
Marks  cf  f 
010  10  10 
1020  50  5010=40 
2030  130  13050=80 
3040  270  270130=140 
4050  440  440270=170 
5060  570  570440=130 
6070  670  670570=100 
7080  740  740670=70 
8090  780  780740=40 
90100  800  800780=20 
Question:11
Form the frequency distribution table from the following data :
Answer:
Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
class  f 
010  3432=2 
1020  3230=2 
2030  3027=3 
3040  2723=4 
4050  2317=6 
5060  1711=6 
6070  116=5 
7080  64=2 
8090  4 
Question:12
Answer:
Solution. Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to
summarize categorical variables.
( because first term of frequency and cumulative frequency is same )
12 + b = 25
b = 25 â€“ 12
b = 13
25 + 10 = c
35= c
c + d = 43
35 + d = 43
d = 43 â€“ 35
d=8
43 + e = 48
e = 48 â€“ 43
e =5
48+2 = f
50 = f
Ans. a = 12, b = 13, c = 35, d = 8, e = 5, f = 50
Question:13
Answer:
Solution. Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
Age (in year)  No.of patients 
less than 10  0 
less than 20  60+0 = 60 
less than 30  42+60 = 102 
less than 40  102+55 =157 
less than 50  157+70 = 227 
less than 60  227+53 =280 
less than 70  280 +20 =300 
Age (in year)  No.of patients 
More than or equal to 10  60+42+55+70+53+20 = 300 
Form the frequency distribution table for the data
Answer:
Solution. Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
Marks  Number of students CF  f 
0 20  17  17 
20 40  22  2217 = 5 
40 60  29  29 22 = 7 
60 80  37  3729 = 8 
80100  50  50 37 = 13 
Question:15
Weekly income of 600 families is tabulated below :
Compute the median income.
Answer:
Answer. [1263.15]
Solution. n = 600
l= 1000, l = 1000, cf = 250, f = 190
median =
Median = 1263.15
Question:16
Calculate the median bowling speed.
Answer:
Answer. [109.16]
Solution. Here n = 33
h = 15, f = 9 , cf = 11
Median
= 100 + 9.16 109.16
Question:17
The monthly income of 100 families are given as below :
Answer:
Answer. [11875]
Solution. Here l = 10000, f_{1} = 41, f_{0} = 26, f_{2} = 16, h = 5000
Mode =
=
=
= 10000 + 1875 = 11875
Modal income is 11875 Rs.
Question:18
The weight of coffee in 70 packets are shown in the following table :
Answer:
Answer. [201.7 g]
Solution.
Here l = 201, f1 = 26, f0 = 12, f2 = 20, h = 1
Question:19
Answer:
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number of cases after thrown of two dice = 36
(i) Same number = (1, 1), (2, 2), (3, 3), (4. 4), (5, 5), (6, 6)
Same number cases = 6
Let A be the event of getting same number.
Probability [p(A)] =
(ii) Different number cases = 36 â€“ same number case
= 36 â€“6 = 30
Let A be the event of getting different number
Probability [p(A)]=
Question:20
Answer:
(i) Answer. [1/6]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases after throwing of two dice = 36
Cases when total is 7 = (1, 6), (6, 1), (3, 4), (4, 3), (2, 5), (5, 2)
Total cases = 6
Let A be the event of getting total 7
Probability [p(A)]=
Probability of getting sum 7 =
(ii) Answer. [5/12]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
Prime number as a sum = (1, 1), (1, 2), (2, 1), (1, 4),
(4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (3, 4), (4, 3), (2, 5), (5, 2), (6, 5), (5, 6)
Cases = 15
Probability =
Probability that sum is a prime number =
(iii) Answer. [0]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
pairs from which we get sum 1 = 0
Cases = 0
Probability =
Probability of getting sum 1 =
Question:21
Answer:
(i) Answer. [1/9]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
For getting product 6 = (1, 6,), (6, 1), (2, 3), (3, 2)
Cases = 4
Probability =
Probability of getting product 6 =
(ii) Answer. [1/9]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
product 12 = (2, 6), (6, 2), (3, 4), (4, 3)
Cases = 4
Probability =
Probability of getting product 12 =
(iii) Answer. [0]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
Product 7 = 0 (case)
Cases = 0
Probability =
Probability of getting product 7 =
Question:22
Answer:
Answer. [4/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases in throwing two dice = 36
Product less than 9 cases = (1, 1). (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (4, 1), (4, 2), (5, 1), (6, 1)
Number of favourable cases = 16
Probability =
Probability of getting product less than 9 =
Question:23
Answer:
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number of cases = 36
case of getting sum 2 = ( 1 , 1 ) ( 1 , 1 )
Probability =
Probability of getting sum 2 =
case of getting sum 3 = (1, 2), (1, 2), (2, 1), (2, 1)
Probability =
Probability of getting sum 3=
case of getting sum 4 = (1, 3), (1, 3), (2, 2), (2, 2), (3, 1), (3, 1)
Probability =
Probability of getting sum 4=
case of getting sum 5 = (2, 3), (2, 3), (4, 1),(4,1) (3, 2), (3, 2)
Probability =
Probability of getting sum 5 =
case of getting sum 6 = (3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)
probability =
Probability of getting sum 6=
case of getting sum 7 = (4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)
probability =
Probability of getting sum 7=
case of getting sum 8 = (5, 3), (5, 3), (6, 2), (6, 2)
probability =
Probability of getting sum 8=
case of getting sum 9 = (6, 3), (6, 3)
probability =
Probability of getting sum 9=
Question:24
A coin is tossed two times. Find the probability of getting at most one head.
Answer:
Answer. [3/4]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 4 (HH, TT, HT, TH)
Cases of at most 1 head = HT, TH, TT
Probability =
Probability of getting at most 1 head =
Question:25
Answer:
(i) Answer. [1/8]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Possible outcomes = (HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
Total cases = 8
Cases of getting all heads = (HHH)
Number of favourable cases = 1
Probability =
Probability of getting all heads =
(ii) Answer.[1/2]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Possible outcomes = 8 (HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
Cases of getting at least 2 heads = (HHH, HHT, HTH, THH)
Favorable cases = 4
Probability =
Probability of getting at least 2 heads =
Question:26
Answer:
Answer. [2/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Cases of getting difference 2 = (1, 3), (3, 1), (2, 4), (4, 2), (3, 5), (5, 3), (4, 6), (6, 4)
Favourable cases = 8
Probability =
Probability of getting difference 2 =
Question:27
Answer:
(i) Answer. [5/11]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = 10 + 5 + 7 = 22
Red balls = 10
Probability =
Probability of getting red ball =
(ii) Answer. [7/22]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = 10 + 5 + 7 = 22
Green balls = 7
Probability =
Probability of getting green ball =
(iii) Answer. [17/22]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = 10 + 5 + 7 = 22
Not a blue ball = 22 â€“ (blue ball)
= 22 â€“ 5 = 17
robability =
Probability of getting not a blue ball =
Question:28
Answer:
(i) Answer. [13/49]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 â€“ 3 ( three cards are removed)
= 49
Total hearts = 13
Favourable cases = 13
Probability =
Probability of getting a heart =
(ii) Answer. [3/49]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 â€“ 3 ( three cards are removed)
= 49
Total king = 4 â€“ 1 = 3 ( 1 king is removed)
favourable cases = 3
bability =
Probability of getting a King=
Question:29
Answer:
(i) Answer. [10/49]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 â€“ 3 = 49 ( three cards are removed)
Total club = 13 â€“ 3 = 10 ( 3 club cards are removed)
favourable cases = 10
Probability =
Probability of getting a club =
(ii) Answer. [1/49]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 â€“ 3 = 49 ( three cards are removed)
10 of heart = 1
favourable cases = 1
Probability =
Probability of getting a heart =
Question:30
Answer:
(i) Answer. [1/10]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 â€“ 12 = 40 ( 12 cards are removed)
card with number 7 = 4
favourable cases = 4
probability =
Probability of getting card 7=
(ii) Answer. [3/10]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 â€“ 12 = 40 ( 12 cards are removed)
Cards greater than 7 =8,9,10 (3 Ã— 4 = 12)
favourable cases = 12
probability =
Probability of getting card 7=
(iii) Answer. [3/5]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 â€“ 12 = 40 ( 12 cards are removed)
Cards less than 7 = 1, 2, 3, 4, 5, 6 (6 Ã— 4 = 24)
favourable cases = 24
probability =
Probability of getting card 7=
Question:31
Answer:
(i) Answer. [14/99]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number = 99 (between 0 to 100)
Number divisible by 7 = (7, 14, 21, 28,35, 42, 49, 56, 63, 70, 77, 84, 91, 98)
Favourable cases = 14
Probability =
Probability of getting number divisible by 7 =
(ii) Answer. [85/99]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number between 0 to 100 = 99
Number divisible by 7 = (7, 14, 21, 28,35, 42, 49, 56, 63, 70, 77, 84, 91, 98)
Favourable cases = 14
robability =
Probability of getting number divisible by 7 =
Question:32
Answer:
(i) Answer. [1/2]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total numbers from 2 to 101 = 100
Total even numbers from 2 to 101 = 50
Favourable cases = 50
Probability =
Probability that card is with even number =
(ii) Answer. [9/100]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number from 2 to 101 = 100
Square numbers from 2 to 101 = (4, 9, 16, 25, 36, 49, 64, 81, 100)
Favourable cases = 9
Probability =
Probability that the card is with a square number =
Question:33
Answer:
Answer. [21/26]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total alphabets = 26
Total consonant = 21
Favourable cases = 21
Probability =
Probability that alphabet is consonant =
Question:34
Answer:
Answer. [0.69]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total envelopes = 1000
Envelopes with no cash prize = Total envelopes â€“ envelopes with cash prize
= 1000 â€“ 10 â€“ 100 â€“ 200 = 690
Favourable cases = 690
Probability =
Probability that the envelope is no cash prize =
Question:35
Answer:
Answer. [11/75]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total slips = 25 + 50 = 75
Slips marked other than 1 = Rs. 5 slips + Rs. 13 slips
= 6 + 5 = 11
Favourable cases = 11
Probability =
Probability that slips is not marked 1
Question:36
Answer:
Answer. [5/23]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total bulbs = 24
Defective = 6
not defective = 18
Probability that the bulb is not defective =
Let the bulbs is defective and it is removed from 24 bulb.
Now bulbs remain = 23
In 23 bulbs, nondefective bulbs = 18
defective = 5
Probability =
Now probability that the bulb is defective = .
Question:37
Answer:
(i) Answer. [4/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 8 + 10 = 18
Total triangles = 8
Favourable cases = 8
Probability =
Probability that piece is a triangle
(ii) Answer. [5/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 8 + 10 = 18
Total square = 10
Favourable cases = 10
Probability =
Probability that the piece is a square =
(iii) Answer. [1/3]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 10 + 8 = 18
Square of blue color = 6
favourable cases = 6
Probability =
Probability that piece is a square of blue color =
(iv) Answer. [5/18]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 10 + 8 = 18
triangle of red color = 8 â€“ 3 = 5
favourable cases = 5
Probability =
Probability that piece is a triangle of red colour
Question:38
Answer:
(i) Answer. [1/8]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
case in which the lose entry = 8 â€“ (in which she gets entry book + in which she gets double)
= 8 â€“ 6 (HHT, HTH, THH, TTH, THT, HTT) â€“ 1(HHH)
= 8 â€“ 7 = 1
Favourable cases = 1
Probability =
Probability that she will lose money =
(ii) Answer. [1/8]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
case in which she gets double entry = HHH
favourable cases = 1
Probability =
Probability that she gets double entry fee =
(iii) Answer. [3/4]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
case in which she gets entry book = 6(HHT, HTH, THH, TTH, THT, HTT)
favourable cases = 6
Probability =
Probability that she gets entry fees =
Question:39
Answer:
(i) Answer. [6]
Solution. Count the number of sums we can notice by using two dice of (0, 1, 1, 1, 6, 6) type.
We can get a sum of 0 = (0,0)
We can get a sum of 1 = (0,1) , (1,0)
We can get a sum of 2 = (1,1)
We can get a sum of 6 = (0,6) , (6,0)
We can get a sum of 7 = (6,1) , (1,6)
We can get a sum of 12 = (6,6)
We can get a score of 0, 1, 2, 6, 7, 12
Hence we can get 6 different scores.
(ii) Answer. [4/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
Case of getting sum 7 = (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (6,1), ( 6,1), (6,1), ( 6,1), (6,1), (6,1),
Number of favourable cases = 12
Probability =
Probability of getting a total 7 =
Question:40
(i) Answer. [7/8]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total mobiles = 48
Minor defective = 3
major defective = 3
good = 42
Varnika buy only good so favourable cases = 42
Probability =
Probability that acceptable to Varnika =
(ii) Answer. [15/16]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
total mobiles = 48
good = 42
minor defect = 3
major defect = 3
trader accept only good and minor defect.
So favourable cases = 42 + 3 = 45
Probability =
Probability that trader accept
Question:41
Answer:
(i) Answer. [5/6]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = red + white + blue
24 = x + 2x + 3x
6x = 24
x = 4
Red balls = x = 4
White balls = 2x = 2 Ã— 4 = 8
Blue balls = 3x = 3 Ã— 4 = 12
Probability =
Probability that ball is not red =
(ii) Answer. [1/3]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = red + white + blue
24 = 6x
x = 4
white balls = 2x = 2 Ã— 4 = 8
Favourable cases = 8
Probability =
Probability of getting on white ball =
Question:42
Answer:
(i) Answer. [0.009]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 1000
Player wins prize with cards = (529, 576, 625, 676, 729, 784, 841, 900, 961)
Favourable cases = 9
Probability =
Probability that player wins =
(ii) Answer. [0.008]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Now the total cards are = 1000 â€“ 1 = 999
Now the total winning cards = 9 â€“ 1 = 8
Probability =
Probability that second player wins after first =
Question:1
Find the mean marks of students for the following distribution
Answer:
Answer. [51.75]
Solution. Here we calculate mean by following the given steps:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
Marks
xi
cf
fi
fixi
010
5
80
8077 = 3
15
1020
15
77
78772 = 5
75
2030
25
72
7265 =7
175
3040
35
65
6555 = 10
350
4050
45
55
5543 = 12
540
5060
55
43
4328 = 15
825
6070
65
28
2816 = 12
780
7080
75
16
1610 =6
450
8090
85
10
108 = 2
170
90100
95
8
80 = 8
760
Question:2
Determine the mean of the following distribution :
Answer:
Answer. [48.4]
Solution. Here we calculate mean by following the given steps:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
Marks
xi
cf
fi
fixi
010
5
5
5
15
1020
15
9
95 =4
75
2030
25
17
179 = 8
175
3040
35
29
2917 = 12
350
4050
45
45
4529 = 16
540
5060
55
60
6045 = 15
825
6070
65
70
7060 = 10
780
7080
75
78
7870 = 8
450
8090
85
83
8378 = 5
170
90100
95
85
8583 = 2
760
Question:3
Find the mean age of 100 residents of a town from the following data :
Answer:
Answer. [31]
Solution. Here we calculate mean by following the given steps:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
Marks
xi
fi
010
5
10090 =10
50
1020
15
9075 = 15
225
2030
25
7550 = 25
625
3040
35
5025 =25
875
4050
45
2515 =10
450
5060
55
155 = 10
550
6070
65
50 = 5
325
Question:4
The weights of tea in 70 packets are shown in the following table :
Find the mean weight of packets.
Answer:
Answer. [201.95]
Solution. Here we calculate mean by following the given steps:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
Now divide sum of (fx) by sum of (f)
Weight
xi
fi
fixi
200201
200.5
13
2606.5
201202
201.5
27
5440.5
202203
202.5
18
3645.0
203204
203.5
10
2035.0
204205
204.5
1
204.5
205206
205.5
1
205.5
Question:5
The weights of tea in 70 packets are shown in the following table :
Draw the less than type ogive for this data and use it to find the median weight.
Answer:
Answer. [201.8]
Solution.
Weight  cf 
Less than 201  13 
Less than 202  27+13=40 
Less than 203  40+18=58 
Less than 204  58+10=68 
Less than 205  68+1 =69 
Less than 206  69+1 = 70 
Hence the median is 201.8
Question:6
The weights of tea in 70 packets are shown in the following table :
Answer:
Answer. [201.8]
Solution.
Less than type  More than type  
Weight  Number of packets  Number of packets  Number of students 
Less than 200  0  More than or equal to 200  70 
Less than 201  13  More than or equal to 201  7013 = 57 
Less than 202  40  More than or equal to 202  5727 =30 
Less than 203  58  More than or equal to 203  3018 =12 
Less than 204  68  More than or equal to 204  1210 = 2 
Less than 205  69  More than or equal to 205  21 = 1 
Less than 206  70  More than or equal to 206  11 = 0 
Hence median = 201.8
Question:7
The table below shows the salaries of 280 persons.
Calculate the median and mode of the data.
Answer:
Solution.
Salary  fi  cf 
510  49  49 
1015  133  49+133=182 
1520  63  182+63=245 
2025  15  245+15 = 260 
2530  6  260+6 = 266 
3035  7  266+7 = 273 
3540  4  273+4 = 277 
4045  2  277+2 = 279 
4550  1  279+1 = 280 
f_{1} = 49, f_{m}= 133, f_{2}= 63, cf = 49, f = 133
l = 10, h = 5
median =
=
=
=
In thousands = 13.421 Ã— 1000 = 13421 Rs.
Mode =
=
=
=10 + 2.727
=12.727
In thousands = 12.727 Ã— 1000 = 12727 Rs.
Question:8
Answer:
Solution.
Class  (f_{i})  x_{i}  f_{i}  
020  17  10  2  34 
2040  f_{1}  30  1  f_{1} 
4060  32  50=a  0  0 
6080  f_{2}  70  1  f_{2} 
80100  19  90  2  38 
Sum of all frequencies = 120
68 + f_{1} + f_{2} = 120
f_{1} + f_{2} = 52 â€¦(1)
a = 50, h = 20
mean =
50= 50 +
0= (4 + f_{2} â€“ f_{1})
â€“f_{2} + f_{1} = 4 â€¦(2)
add (1) and (2) we get
2f_{1} = 56
Put f_{1} = 28 in equation (1)
f_{2} = 52 â€“ 28
Question:9
Answer:
Solution.
marks  Frequency  Cummulative frequency 
2030  1  p 
3040  15  15+p 
4050  25  40+p = cf 
5060  20=f  60+p 
6070  q  68+p+q 
7080  8  68+p+q 
8090  10  78+p+q 
n = 90,
l = 50, f = 20, cf = 40 + p, h = 10
median =
5 â€“ p = 0
p = 5
78 + 5 + q = 90
q = 90 â€“ 83
q = 7
Question:10
The distribution of heights (in cm) of 96 children is given below :
Answer:
Answer. [139]
Solution.
Height  Number of children 
less than 124  0 
Hence the median is = 139
Question:11
Size of agricultural holdings in a survey of 200 families is given in the following table:
Compute median and mode size of the holdings
Answer:
Answer. [17.77]
Solution.
Size of agricultural holdings  fi  cf 
05  10  10 
510  15  25 
1015  30  55 
1520  80  135 
2025  40  175 
2530  20  195 
3035  5  200 
(i) Here n = 200
which lies in interval (15 â€“ 20)
l = 15, h = 5, f = 80 and cf = 55
=
l = 15, f_{m} = 80, f_{1} = 30, f_{2} = 40 and h = 5
=
=15 + 2.77 = 17.77
Question:12
The annual rainfall record of a city for 66 days is given in the following table.
Calculate the median rainfall using ogives (of more than type and of less than type)
Answer:
Answer. [20]
Solution.
(i) less than type  (ii) more than type  
Rain fall  No.of days  Rain fall  Number of days 
less than 0  0  more than or equal to 0  66 
less than 10  0+22 = 22  more than or equal to 10  6622 = 44 
less than 20  22+10 = 32  more than or equal to 20  4410 = 34 
less than 30  32+8 = 40  more than or equal to 30  348 = 26 
less than 40  40+15 = 55  more than or equal to 40  2615 = 11 
less than 50  55+5 =60  more than or equal to 50  11  5 =6 
less than 60  60+6 =66  more than or equal to 60  66 =0 
Now let us draw ogives of more than type and of less than type then find the median
Here median is 20
Question:13
The following is the frequency distribution of duration for100 calls made on a mobile phone :
Answer:
Answer. [170]
Solution.
Duration  fi  xi  f_{i}u_{i}  
95125  14  110  2  28 
125155  22  140  1  22 
155=185  28  170 = a  0  0 
185215  21  200  1  21 
215245  21  230  2  30 
a = 170, h = 30
Average =
less than type  
Duration  Number of calls 
less than 95  0 
n = 100
median is 170
Question:14
Answer:
(i) Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
Distance  f_{i}  CF 
020  6  6 
2040  11  6+11 = 17 
4060  17  17+17 = 34 
6080  12  34 + 12 = 46 
80100  4  46 + 4 =50 
(ii) Answer. [49.41]
Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables
Distance  Cumulative frequency (C.F) 
0  0 
n = 50
median = 49.41
(iii)Answer. [49.41]
Solution. n = 50
which lies in interval 40 â€“ 60
l = 40, h = 20, CF = 17 and f = 17
=
=
=40 + 9.41
= 49.41
(iv) Yes, the median distance calculated in (ii) and (iii) are same.
NCERT Exemplar Solutions Class 10 Maths Chapter 13 Important Topics:
NCERT exemplar class 10 maths solutions chapter 13 deals with a wide range of concepts mentioned below:
â—Š How to find out the central tendency is of any given data.
â—Š Methods to find out the mean of any group data.
â—Š The direct method, step deviation method and assumed mean method to find out the mean of any group data.
â—Š NCERT Exemplar class 10 maths solutions chapter 13 discusses cumulative frequency distribution and its graphical representation.
â—Š How to find out the mode and median of the given data.
â—Š Different techniques to find out medians.
â—Š Conditional probabilities, independent events, and Bayes theorem.
NCERT Class 10 Solutions for Other Subjects:
NCERT Exemplar Class 10 Maths solutions
NCERT Exemplar Class 10 Science solutions
NCERT Class 10 Maths Exemplar Solutions for Other Chapters:
Chapter 
NCERT exemplar solution for class 10 Maths Chapter 1 Real Numbers 
NCERT exemplar solutions for class 10 Maths Chapter 2 Polynomials 
NCERT exemplar solution for class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 
NCERT exemplar solution for class 10 Maths Chapter 4 Quadratic Equations 
NCERT exemplar solution for class 10 Maths Chapter 5 Arithmetic Progressions 
NCERT exemplar solution for class 10 Maths Chapter 6 Triangles 
NCERT exemplar solution for class 10 Maths Chapter 7 Coordinate Geometry 
NCERT exemplar solution for class 10 Maths Chapter 8 Introduction to Trigonometry & Its Equations 
NCERT exemplar solution for class 10 Maths Chapter 9 Circles 
NCERT exemplar solution for class 10 Maths Chapter 10 Constructions 
NCERT exemplar solution for class 10 Maths Chapter 11 Areas related to Circles 
NCERT exemplar solution for class 10 Maths Chapter 12 Surface Areas and Volumes 
NCERT exemplar solution for class 10 Maths Chapter 13 Statistics and Probability 
Features of NCERT exemplar class 10 maths solutions chapter 13:
These class 10 maths NCERT exemplar chapter 13 solutions emphasise the methods to find out mean, median, and mode. In this chapter, students will understand the experimental and statistical approach of probability. Students will learn the condition for multiplying probability to find out the probability of any composite event. Statistics and Probability based practice problems can be easily studied and practiced using these class 10 maths NCERT exemplar solutions chapter 13 Statistics and Probability. The students can comfortably sail through the NCERT class 10 maths, RD Sharma class 10 maths, A textbook for Mathematics by Monica Kapoor, and RS Aggarwal class 10 maths et cetera.
Check Chapterwise Solutions of Book Questions
Chapter No.  Chapter Name 
Chapter 1  
Chapter 2  
Chapter 3  NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables 
Chapter 4  NCERT solutions for class 10 maths chapter 4 Quadratic Equations 
Chapter 5  NCERT solutions for class 10 chapter 5 Arithmetic Progressions 
Chapter 6  
Chapter 7  NCERT solutions for class 10 maths chapter 7 Coordinate Geometry 
Chapter 8  NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 
Chapter 9  NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry 
Chapter 10  
Chapter 11  
Chapter 12  NCERT solutions for class 10 chapter maths chapter 12 Areas Related to Circles 
Chapter 13  NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes 
Chapter 14  
Chapter 15 
Frequently Asked Question (FAQs)  NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability
Question: What is the probability that grass in the garden is green?
Answer:
This is incomplete information about the grass. We cannot find its probability due to data insufficiency. Generally, people mistakenly assume that the probability of grass to be green is half. But being green or not agree or not equally likely.
Question: What does central tendency of the data mean?
Answer:
For any data central tendency implies mean median and mode.
Question: Is the chapter Statistics and Probability important for Board examinations?
Answer:
The chapter Statistics and Probability is important for Board examinations as it holds around 810% weightage of the whole paper.
Question: Are these solutions accessible offline?
Answer:
Yes, the NCERT exemplar class 10 maths solutions chapter 13 pdf download feature provided this solution for students practicing NCERT exemplar Class 10 maths chapter 13.
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