Statistics(10th Grade > Mathematics ) Questions and answers for exam Preparation

10th Grade > Mathematics

STATISTICS MCQs

Total Questions : 60 | Page 2 of 6 pages

\sum

fx = 5x + 2,

\sum

f = 12. If the mean of the distribution is 6, what is the value of x?

Discuss Question

Answer: Option B. -> 14
:
B
We know Mean =

\frac{\sum f x}{\sum f}

Hence,

\frac{5 x + 2}{12}

= 6

\Rightarrow

5x + 2 = 12

\times

\Rightarrow

5x = 72 – 2 = 70

\Rightarrow

x = 14

Question 12. 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

\begin{matrix} x_{i} & f_{i} & f_{i} x_{i} 10 & 2 & 20 13 & 5 & 65 16 & 7 & 112 19 & 6 & 114 \sum f_{i} = 20 & \sum f_{i} x_{i} = 311 \end{matrix}

Mean

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

= \frac{311}{20}

= 15.55

Question 13. In the fromula for mode of a grouped data ,
mode

= l + [\frac{t_{1} - t_{0}}{2 f_{1} - f_{0} - f_{2}}] \times h

,
where symbols have their usual meaning

f_{1}

represents

Frequency of Modal class
Frequency of the class preceding the modal class
Frequency of the class succeeding the modal class
Frequency of median class

Discuss Question

Answer: Option A. -> Frequency of Modal class
:
A
In the formula for mode,

f_{1}

represents the frequency of the modal class.

Question 14. Class mark of a class is _____

Upper limit + lower limit
upper limit + lower limit2
upper limit - lower limit2
upper limit - lower limit

Discuss Question

Answer: Option B. -> upper limit + lower limit2
:
B
Class mark is the midpoint of a class interval.
Therefore, its formula is given by

\frac{upperlimit + lowerlimit}{2}

Question 15. Find the mode of the following data.

\begin{matrix} C l a s s i n t e r v a l & 0 - 100 & 100 - 200 & 200 - 300 & 300 - 400 & 400 - 500 F r e q u e n c y & 7 & 11 & 15 & 9 & 8 \end{matrix}

___

Discuss Question

:
Modal Class is 200-300
Mode=I+

(\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}})

\times

h
=200+

\frac{15 - 11}{30 - 11 - 9}

\times

100=240

Question 16. The below table shows that profit made by a group of shops in mall. Then the median is _____.

\begin{matrix} P r o f i t p e r s h o p l e s s t h a n (%) & 10 & 20 & 30 & 40 & 50 & 60 N o . o f s h o p s & 12 & 30 & 57 & 77 & 94 & 100 \end{matrix}

27.4
31.2
26.8
23.7

Discuss Question

Answer: Option A. -> 27.4
:
A
Firstly, the cumulative frequency (CF) table is drawn.

\begin{matrix} C . I & C F & F r e q u e n c y 0 - 10 & 12 & 12 10 - 20 & 30 & 18 20 - 30 & 57 & 27 30 - 40 & 77 & 20 40 - 50 & 94 & 17 50 - 60 & 100 & 6 T o t a l & 100 \end{matrix}

\frac{n}{2} = \frac{100}{2} = 50

C.F nearest to 50 and greater than 50 is 57.

∴ M e d i a n = l + (\frac{\frac{n}{2} - c f}{f}) \times h

Where

l

is lower class limit of median class.

n

is total number of observations.

c f

is the cumulative frequency of the class preceding the median class.

f

is the frequency of the median class and h is the class size.

M e d i a n = 20 + (\frac{50 - 30}{27}) \times 10

Median = 27.4

Question 17. The following table gives the life time of 200 neon lamps. Find the median class.

2500 - 3000
1000 - 1500
1500 - 2000
3000 - 3500

Discuss Question

Answer: Option A. -> 2500 - 3000
:
A

The median class is identified as the class whose cumulative frequency will be greater than

\frac{n}{2}

, where 'n' is the sum of the frequencies. (n=200).
Next, we divide 200 by 2.

\frac{n}{2} = \frac{200}{2} = 100

We now locate the class whose cumulative frequency is greater than 100.
"2500 - 3000" is the class whose cumulative frequency 155 is greater than 100.

∴

The median class is 2500 - 3000.

Question 18. The following data shows monthly savings of 100 families . Calculate the mode of the given frequency distribution.

\begin{matrix} M o n t h l y s a v i n g s (R s) & N u m b e r o f f a m i l i e s 1000 - 2000 & 14 2000 - 3000 & 15 3000 - 4000 & 21 4000 - 5000 & 27 5000 - 6000 & 25 \end{matrix}

4790
4760
4750
4780

Discuss Question

Answer: Option C. -> 4750
:
C
Modal class is 4000-5000 within

f_{1}

= 27,

f_{0}

= 21,

f_{2}

= 25, I = 4000, h = 1000
Mode = l +

\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times h

Mode = 4000 +

\frac{6}{8}

\times

1000
= 4750

Question 19.

1 - 2
2 - 3
6 - 7
7 - 8

Discuss Question

Answer: Option B. -> 2 - 3
:
B
Here we locate a class with the maximum frequency. From the given table we find the maximum frequency as 12 whose class is 2 - 3. Therefore the modal class is 2 - 3.

Question 20. For which data set, the mean is not a good representative measures of central tendency.

There are 7 classes with the below frequencies. {10, 12, 15, 17, 14, 15, 17}
There are 6 classes with the below frequencies. {2, 20, 25, 17, 19, 22}
There are 7 classes with the below frequencies. {5, 7, 5, 6, 4, 5, 7}
There are 6 classes with the below frequencies. {17, 20, 25, 17, 19, 22}

Discuss Question

Answer: Option B. -> There are 6 classes with the below frequencies. {2, 20, 25, 17, 19, 22}
:
B
We know that extreme values in the data affect the mean. Here only for the data set{2, 20, 25, 17, 19, 22}, we find the extreme value as one class has the frequency as 2 and the others have frequency 20, 25, 17, 19 and 22.