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

10th Grade > Mathematics

STATISTICS MCQs

Total Questions : 60 | Page 4 of 6 pages

Question 31.

The median for grouped data is formed by using the formula:

$l + (\frac{\frac{n}{2} - c_{f}}{f}$ ) $\times h$
$l - (\frac{\frac{n}{2} - c f}{f}$ ) $\times h$ .
$l + (\frac{\frac{n}{2} + c f}{f}$ ) $\times h$ .
$l + \frac{2}{n}$

Discuss Question

Answer: Option A. ->

l + (\frac{\frac{n}{2} - c_{f}}{f}

)

\times h

:
A

The median for grouped data is formed by $l + (\frac{\frac{n}{2} - c_{f}}{f}$ ) $\times h$ .

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.

Question 32.

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 33.

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

Answer: Option C. -> 4750
:

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 34.

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 35.

In the assumed mean method, if A is the assumed mean, then deviation $d_{i}$ is :

$x_{i} - A$
$x_{i} + A$
$x_{i}$
$A - x_{i}$

Discuss Question

Answer: Option A. ->

x_{i} - A

:
A
The deviation is

d_{i} = x_{i} - A

Question 36.

The times (in seconds) taken by 150 athelets to run a 110 m hurdle race are tabulated below

$\begin{matrix} Class & 13.8 - 14 & 14 - 14.2 & 14.2 - 14.4 & 14.4 - 14.6 & 14.6 - 14.8 & 14.8 - 15 Frequency & 2 & 4 & 5 & 71 & 48 & 20 \end{matrix}$

The number of atheletes who completed the race in less than 14.6 s is

Discuss Question

Answer: Option C. -> 82
:
C

The number of atheletes who completed the race in less than 14.6

= 2 + 4 + 5 + 71

= 82

Question 37.

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 38.

In the assumed mean method, if A is the assumed mean, then deviation $d_{i}$ is :

$x_{i} - A$
$x_{i} + A$
$x_{i}$
$A - x_{i}$

Discuss Question

Answer: Option A. ->

x_{i} - A

:
A

In statistics, 'the assumed mean' is a method for calculating the arithmetic mean and standard deviation of a data set. It simplifies calculating accurate values by hand.

During the application of the short-cut method for finding the mean, the deviation d, are divisible by a common number ‘h’. In this case, the di = xi – A is reduced to a great extent as 'di' becomes $\frac{d i}{h}$ . In this method, we divide the deviations by the same number to simplify calculations. The deviation is $d_{i} = x_{i} - A$

Question 39.

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 Following Table Gives The Life Time Of 200 Neon Lamps. F...

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 40.

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.