Exercise 13.1
Multiple Choice Questions:

(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. (c)          
Explanation:
Here, a is assumed mean from class marks (xi) and d_i is the difference between a and each of the x_i’s (mid-value).
Thus, d_i’s are the deviation from a of mid-points 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 class
(d) centred at the lower limits of the class
Answer. (b)          
Expalnation:
While grouping the data all the observations between lower and upper limits of class marks are taken in one group then mid value is taken for any further calculation. In computing the mean of grouped data, it is assumed that the frequency of each class interval is centred around its mid value or class mark.


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 of these
Answer. (b)
Explanation:
Since, the intersection point of less than ogive and more than ogive gives the median on the abscissa because on X-axis we take the upper or lower limits respectively and on Y-axis we take cumulative frequency.

Answer. (b)
Explanation:
Commulative frequency for the given data is given below:
Class
Frequency
Cumulative frequency
0-5
10
10
5-10
15
25
10-15
12
37
15-20
20
57
20-25
9
66
Now, the modal class is the class having the maximum frequency.
The maximum frequency 20 belongs to class (15 – 20).
So, lower limit of modal class is 15.
And, the class whose cumulative frequency is greater than (and nearest to) n/2 is called the median class.
Here, n/2 = 66/2 = 33, which lies in the interval 10-15.
So, lower limit of the median class is 10.
Hence, required sum of lower limits of the median class and modal class is 10 + 15 = 25.

(a) 17                                    
(b) 17.5                                 
(c) 18                                     
(d) 18.5
Answer. (b)
Explanation:
First make the classes continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each class.
Then find the Commulative frequency for the given data as below:
Class
Frequency
Cumulative frequency
0.5-5.5
13
13
5.5-11.5
10
23
11.5-17.5
15
38
17.5-23.5
8
46
23.5-29.5
11
57
Now, the class whose cumulative frequency is greater than (and nearest to) n/2 is called the median class.
Here, n/2 = 57/2 = 28.5 which lies in the interval 11.5 – 17.5.
Hence, the upper limit is 17.5.
Question. 8 For the following distribution,
Marks
Number of students
Below 10
3
Below 20
12
Below 30
27
Below 40
57
Below 50
75
Below 60
80
the modal class is
(a) 10-20                                     
(b) 20-30                                     
(c) 30-40                                     
(d) 50-60
Answer. (c)
Explanation:
Commulative frequency for the given data is calculated as below:
Marks
Number of students
Cumulative frequency
Below 10
 3
3
10-20
 9
12
20-30
15
27
30-40
30
57
40-50
18
75
50-60
5
80
Now the class whose frequency is maximum is called the modal class. Here, we see the highest frequency is 30, which lies in the interval 30-40. Therefore, 30-40 is the modal class.
Question.  9 Consider the data

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.  (c)
Explanation:
Commulative frequency for the given data is calculated as below:
Class
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
7
63
185-205
4
67
Now, the class whose cumulative frequency is greater than (and nearest to) n/2 is called the median class.
Here, n/2 = 67/2 = 33.5, which lies in the interval 125 – 145.
Hence, upper limit of median class is 145.
Again, the class whose frequency is maximum is called the modal class.
Here, 20 is the highest frequency which lies in the interval 125 –145.
Hence, the lower limit of modal class is 125.
Thus, the required difference = Upper limit of median class – Lower limit of modal class
                                                = 145 –125 = 20.
Question. 10 The times (in seconds) taken by 150 atheletes to run a 110 m hurdle race are tabulated below

The number of atheletes who completed the race in less than 14.6 s is
(a) 11    
(b) 71    
(c) 82     
(d) 130
Answer. (c)
Exlanation:
 The number of atheletes who completed the race in less than 14.6
= 2 + 4 + 5 + 71 = 82.
Question.  11 Consider the following distribution
Marks obtained
Number of students
More than or equal to 0
63
More than or equal to 10
58
More than or equal to 20
55
More than or equal to 30
51
More than or equal to 40
48
More than or equal to 50
42
the frequency of the class 30-40 is
(a) 3       
(b) 4       
(c) 48     
(d) 51
Answer. (a)
Explanation:
The given frequency data can be represented in the form of class intervals as follows:
Marks obtained
Number of students
Frequency
0-10
(63 – 58) = 5
5
10-20
(58 – 55) = 3
3
20-30
(55 – 51) = 4
4
30-40
(51 – 48) = 3
3
40-50
(48 – 42) = 6
6
50…
42 − 0 = 42
42
Hence the frequency of 30 – 40 class interval is 3.
Question. 12   If an event cannot occur, then its probability is
(a) 1                        (b) 3/4                        (c) 1/2                (d) 0
Answer. (d)   
Explanation:
The event which cannot occur is called the impossible event with a probability equal to zero.
Question. 13   Which of the following cannot be the probability of an event?
(a) 1/3                     (b) 0.1                  (c) 3                       (d) 17/16
Answer. (d)                   
Explanation:
Probability of an event always lies between 0 and 1.
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. (a)
Explanation:
The probability of an event which is very unlikely to happen is closest to zero and from the given options 0.0001 is closest to zero.
Question. 15   If the probability of an event is P, then the probability of its complementary event will be
(a) P – 1                      (b) P                  (c) 1 – P                  (d) 1 – 1/P
Answer. (c)   
Explanation:
Since, probability of an event + probability of its complementary event = 1.
∴ Probability of its complementary event = 1 – Probability of an event = 1 – 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. (b)
Explanation:
The probability expressed as a percentage always lies between 0 to 100. So, it cannot be less than 0.
Question. 17   lf P(A) denotes the probability of an event A, then
(a) P(A) < 0          (b) P(A) > 1        (c) 0 ≤ P(A) ≤ 1           (d) –1 ≤ P(A) ≤ 1
Answer. (c)   
Explanation:
The probability of an event always lies between 0 and 1.
Question. 18   If a card is selected from a deck of 52 cards, then the probability of its being a red face card is
(a) 3/26                    (b) 3/13                 (c) 2/13                    (d) 1/2
Answer. (a)   
Explanation:
In a deck of 52 cards, there are 12 face cards with 6 red and 6 black cards.
So, probability of getting a red face card = 6/52 = 3/26
Question. 19   The probability that a non-leap year selected at random will contain 53 Sundays is
(a) 1/7                           (b) 2/7                    (c) 3/7                    (d) 5/7
Answer. (a)
Explanation:
A non-leap year has 365 days with 52 weeks and 1 odd day. This 1 odd day may be any day of the week.
Thus, out of 7 days, there is 1 favourable event that this 1 day is Sunday.
Hence the probability of getting 53 Sundays = 1/7
Question. 1 The median of an ungrouped data and the median calculated when the same data grouped are always the same. Do you think that this is a correct statement? Give reason.
Solution.
No,because when we calculate the median of a grouped data, the formula we used is based on the assumption that the observations in the classes are uniformly distributed.
Question. 2 In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula,

Where, a is the assumed mean, a must be one of the mid-point of the classes. Is the last statement correct? Justify your answer.
Solution.
No,the assumed mean must not always be the mid-point of the class interval. It is considered as any value which may simplify the calculation.
Question. 3 Is it true to say that the mean, mode and median of grouped data will always be different? Justify your answer.
Solution.
Yes, the mean, mode and median of grouped data will always be different,
As mean = sum of all observations/ number of observations

And mode = The maximum frequency of a particular observation becomes the mode of that particular data
Hence the three measures will always be different.
Question. 4 Will the median class and modal class of grouped data always be different? Justify your answer.
Solution.
No, the median and modal class of grouped data can also be same in some cases.
Justification:

Here we have,
N = 50  
Then, N/2 =50/2 = 25
The cumulative frequency just greater than N/2 is 26, then the median class is 120 – 140.
Also the maximum frequency here is 14, then the corresponding class 120 – 140 is the modal class.
Hence the median class and modal class are the same.
Q. Two dice are thrown at the same time. Find the probability of getting
(i) same number on both dice.
(ii) different number on both dice.
Sol.
As the two dice are thrown at the same time,                                
So, total number of possible outcomes = 6² = 36
(i) In case of same numbers on both dice the possible outcomes are
      (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).
∴ Number of possible outcomes = 6
So, required probability = 6/36 = 1/6
(ii) In case of different numbers on both dice the possible outcomes are
      = 36 – Number of possible outcomes for same number on both dice
      = 36 – 6 = 30
∴ Required probability = 30/36 = 5/6
Q. Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is
(i) 7?                           
(ii) a prime number ?  
(iii) 1?
Sol.
As the two dice are thrown at the same time,                                
So, total number of possible outcomes = 6² = 36
(i) Sum of the numbers appearing on the dice is 7.
Here, the possible outcomes are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6,1),
Thus, the number of possible outcomes = 6
∴  Required probability = 6/36 = 1/6.
(ii) Sum of the numbers appearing on the dice is a prime number i.e., 2, 3, 5, 7 and 11.
Here, the possible ways are (1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3,4), (4, 3), (5, 2), (6, 1), (5, 6) and (6, 5).
Thus, number of possible outcomes = 15
\ Required probability = 15/36 = 5/12.
(iii) Sum of the numbers appearing on the dice is 1.
It is an impossible event, so its probability is zero.
Q. Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is
(i) 6                            
(ii) 12                         
(iii) 7
Sol.
As the two dice are thrown at the same time,                                
So, total number of possible outcomes = 6² = 36
(i) Product of the numbers on the top of the dice is 6.
Here, the possible ways are (1, 6), (2, 3), (3, 2) and (6, 1).
Thus, the number of possible outcomes = 4
∴ Required probability = 4/36 = 1/9.
(ii) Product of the numbers on the top of the dice is 12.
Here, the possible ways are (2, 6), (3, 4), (4, 3) and (6, 2).
Thus, the number of possible outcomes = 4
∴ Required probability = 4/36 = 1/9.
(iii) Product of the numbers on the top of the dice is 7.
It is an impossible event.
So, its probability is zero.
Q. Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9.
Sol.
As the two dice are thrown at the same time,
Number of total outcomes = 36
For the product of numbers appearing on them to be less than 9, the possible outcomes are
(1, 6), (1,5) (1,4), (1,3), (1,2), (1, 1),(2, 2), (2, 3), (2, 4), (3,2), (4, 2), (4, 1), (3,1), (5,1), (6, 1) and (2,1).
Thus, the number of possible outcomes = 16
∴ Required probability = 16/36 = 4/9
Q. Determine the mean of the following distribution
Marks
Number of students
Below 10
5
Below 20
9
Below 30
17
Below 40
29
Below 50
45
Below 60
60
Below 70
70
Below 80
78
Below 90
83
Below 100
85
Solution.
Given data is of less than type. First we have to convert it to the continuous type.
Here, 5 students have scored marks below 10, i.e. it lies between class interval 0-10 and 9 students have scored marks below 20. So, marks of (9 – 5) = 4 students lie in the class interval 10-20. Proceeding in the same way, we get frequency of all class intervals as below:
Marks
Number of students (f_i)
0-10
5
10-20
9 – 5 = 4
20-30
17 – 9 = 8
30-40
29 – 17 = 12
40-50
45 – 29 = 16
50-60
60 – 45 = 45
60-70
70 – 60 = 10
70-80
78 – 70 = 8
80-90
83 – 78 = 5
90-100
85 – 83 = 2
The complete frequency distribution table for given data can be obtained as follows:

Here, assumed mean, a = 45
And class width, h = 10
Thus, by step deviation method,
Q. The annual rainfall record of a city for 66 days is given in the following table.

Calculate the median rainfall using ogives (or more than type and of less than type)
Solution.
We observe that, number of days for which the rainfall was less than 0 cm is 0. Similarly, the number of days for which the annual rainfall was less than 10 cm is equal to the days having less than 0 cm rainfall plus the days having rainfall from 0-10 cm.
So, the number of days having rainfall less than 10 cm = 0 + 22 = 22 days.
Proceeding in the similar manner, we will get a less than type distribution as follows:
Less than type
Rainfall (in cm)
Number of days
Less than 0
0
Less than 10
0 + 22 = 22
Less than 20
22 + 10 = 32
Less than 30
32 + 8 = 40
Less than 40
40 + 15 = 55
Less than 50
55 + 5 = 60
Less than 60
60 + 6 = 66
Also, we observe that the rainfall record of a city for 66 days is more than or equal to 0 cm. Since, 22 days lies in the interval 0-10. So, annual rainfall record for 66-22 days is more than or equal to 10 cm. Proceeding in the similar manner we will get the more than type distribution as follows:
More than type
Rainfall (in cm)
Number of days
More than or equal to 0
66
More than or equal to 10
66 – 22 = 44
More than orequal to 20
44 – 10 = 34
More than or equal to 30
34 – 8 = 26
More than or equal to 40
26 – 15 =11
More than or equal to 50
11 – 5 = 6
More than or equal to 60
6 – 6 = 0
To draw less than type ogive we plot the points (0, 0), (10, 22), (20, 32), (30, 40), (40, 55), (50, 60), (60, 66) on the paper and join them by free hand.
Also to draw the more than type we plot the points (0, 60), (10, 44), (20, 34), (30, 26), (40, 11), (50, 6) and (60, 0) on the paper and join them by free hand.

The two ogives intersect at a point. From this point of intersection, we draw a line perpendicular to the X-axis meeting the X-axis at point (21.25, 0) which is the required median.
Hence, the median rainfall = 21.25 cm.