Statistics(9th Grade > Mathematics ) Questions and answers for exam Preparation

9th Grade > Mathematics

STATISTICS MCQs

Total Questions : 53 | Page 2 of 6 pages

Question 11. A histogram can never have gaps between two bars.

True
False
22
18

Discuss Question

Answer: Option B. -> False
:
B
A histogram can have a gap between two bars if the intermediate class has 0 frequency.

Question 12. The graph below shows the ages of presidents of a country at the time of inauguration or start of their tenure.

Find the correct option(s) among the following.

The total number of presidents is 45.
The youngest president was 44 years old
The youngest president was 39 years old.
The oldest president is 74 years old.

Discuss Question

Answer: Option B. -> The youngest president was 44 years old
:
B
On summing up all the frequencies, we get 4+11+14+9+4+2=44. Hence the total number of presidents is 44 and option A is incorrect. The graph starts from 39 but the corresponding frequency is 0. Hence, the age of 44 is the youngest age of president. Thus, B is correct and C is incorrect. The last point on thegraph having non-zero frequency is 69. 74 has zero frequency and hence option D is incorrect.

Question 13. Find the range of the following data: 26, 21, 33, 17, 14, 15, 27.

Discuss Question

Answer: Option D. -> 19
:
D
The difference betweenthe highest and lowest values in the data is called the 'range' of the data. So, here the maximum value is 33 and the minimum is 14. So the difference between them i.e. 19 is the range of the given data.

Question 14. What is the median of the following data: 1, 4, 7, 3, 9, 5, 3, 2, 7, 5.

Discuss Question

Answer: Option B. -> 4.5
:
B
For finding the median, the data should be arranged in ascending or descending order first.
Arranging the data in ascending order ,we get :
1, 2, 3, 3, 4, 5, 5, 7, 7, 9.
There are even number of observations.
Hence themedian is the mean value of

(\frac{n}{2})^{t h}

and

[(\frac{n}{2}) + 1]^{t h}

term.
Here the

{\frac{10}{2}}^{t h} t e r m

= 5

^{t h}

term = 4

[(\frac{n}{2}) + 1]^{t h}

term. = 6

^{t h}

term = 5
i.e.

(\frac{4 + 5}{2}) = 4.5

Hence, median of the given data

= 4.5

Question 15. The mean of 15 observations is 8. If the mean of the first eight observations is 7 and that of the last eight observations is 10, then the value of the 8th observation is _____.

Discuss Question

Answer: Option A. -> 16
:
A
We know that,
Mean =

\frac{S u m o f a l l o b s e r v a t i o n s}{N u m b e r o f o b s e r v a t i o n s}

Sum of values of all the observations

= M e a n \times N u m b e r o f o b s e r v a t i o n

= 15 \times 8

= 120

So, sum of thefirst eight observations

= M e a n \times N u m b e r o f o b s e r v a t i o n

= 7 \times 8

= 56

Sum of the last eight observations

= M e a n \times N u m b e r o f o b s e r v a t i o n

= 10 \times 8

= 80

Hence, value of the

8^{t h}

observation

= (80 + 56) - 120

= 16

Question 16. The mode of the given data 7, 9, 11, 13, 9 13, 9, 9, 7, 8 is
___

Discuss Question

:
Mode is that value of the observation which occurs most frequently i.e. an observation with maximum frequency is called mode. In the above set of observations, '9' occurs most frequently and so it is the mode of the given data.

Question 17. The mean of the following distribution is

\begin{matrix} x_{i} & 10 & 13 & 16 & 19 f_{i} & 2 & 5 & 7 & 6 \end{matrix}

15.55
15.35
14.3
16

Discuss Question

Answer: Option A. -> 15.55
:
A
Mean of ungrouped frequency distribution is given by

¯ x = \frac{\sum_{i = 1}^{n} f_{i} x_{i}}{\sum_{i = 1}^{n} f_{i}}

= \frac{(10 \times 2 + 13 \times 5 + 16 \times 7 + 19 \times 6)}{(2 + 5 + 7 + 6)}

= \frac{311}{20}

Therefore, themean of the data is 15.55.

Question 18. The mean of the following distribution is:

\begin{matrix} x_{i} & 10 & 13 & 16 & 19 f_{i} & 2 & 5 & 7 & 6 \end{matrix}

15.2
15.35
15.55
16

Discuss Question

Answer: Option C. -> 15.55
:
C
We know that,
Average or mean value of a given number of observations is the sumof all the observations divided by the total number of observations.

\Rightarrow Mean = \frac{\sum x_{i} f_{i}}{\sum f_{i}}

Now,

\sum x_{i}

f_{i} = 20 + 65 + 112 + 114 = 311

\sum f_{i} = 20

So,

Mean = \frac{\sum x_{i} f_{i}}{\sum f_{i}} = \frac{311}{20} = 15.55

Question 19. The table below shows the sale of shirts of various sizes in a week.

\begin{matrix} Shirt size & 38 & 39 & 40 & 41 & 42 Number sold & 60 & 73 & 89 & 90 & 23 \end{matrix}

The size which is demanded the most is
___.

Discuss Question

:
The frequencies of different observations are given in the table.
It can be seen that shirts of size 41 have the maximum frequency of 90.
So, the mode is 41 and size 41 is the size most in demand.

Question 20. The frequency polygon below shows marks obtained by students of a class.

From the graph, it can be said that the maximum number of students have scored marks between ____.

55-65
50-60
60-70
60

Discuss Question

Answer: Option A. -> 55-65
:
A
The peak of the frequency polygon corresponds to 60 marks. 60 is the class mark of class 55-65. Hence the class 55-65 has the maximum number of students.