Statistics Test

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Statistics Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

The blood groups of 30 students are recorded as follows:
$$A, B, O, A, AB, O, A, O, B, A, O, B, A, AB, B, A,$$
$$ AB, B, A, A, O, A, AB, B, A, O, B, A, B, A$$
What are the frequency distribution for the data $$A$$ and $$AB$$ ?

A
$$10, 2$$

B
$$11, 2$$

C
$$12, 4$$

D
$$13, 5$$

Solution

Frequency is the number of times the data has occured in the data set.

The frequency of the data $$A$$ is $$12$$. Since, $$A$$ has occured $$12$$ times in the given data set.

Similarly $$AB$$ has occured $$4$$ times in the data set. Therefore the frequency of $$AB$$ is $$4$$
Question 2
1 / -0

In the class intervals $$10-20, 20-30$$, the number $$20$$ is included in:

A
$$10-20$$

B
$$20-30$$

C
Both the intervals

D
None of these intervals

Solution

The number is always included in the lower limit of the class interval. Hence, $$20$$ is included in $$20-30$$
Question 3
1 / -0

Let $$m$$ be the mid-point and $$l$$ be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is:

A
$$2m + l$$

B
$$2m - l$$

C
$$m + l$$

D
$$m + 2l$$

Solution

Let $$a$$ be lower class limit, $$l$$ is the upper class limit and $$m$$ is the mid-point
then, mid-point $$m = \dfrac{a +l}{2}$$
Therefore, $$a = 2m - l$$
Question 4
1 / -0

In a frequency distribution, the mid value of a class is $$10$$ and the width of the class is $$6$$. The lower limit of the class is:

A
$$6$$

B
$$7$$

C
$$8$$

D
$$12$$

Solution

$$\textbf{Step1: Find the values}$$
$$\text{Let $$x$$ and $$y$$ be the upper and lower class limit in a frequency distribution.}$$
$$\text{Given, mid value of a class is 10}$$
$$\Rightarrow \dfrac{x+y}{2}=10$$
$$\Rightarrow x+y=20 \ldots (1)$$
$$\text{Also given that, width of the class is 6}$$
$$x-y=6 \ldots(2)$$
$$\text{On Adding equation $$(1)$$ and $$(2)$$, we get}$$
$$\Rightarrow 2 x=20+6 $$
$$\Rightarrow 2 x=26 $$
$$\Rightarrow x=13 $$
$$\text{On putting $$x=13$$ in (1) we get}$$
$$\Rightarrow 13+y=20 $$
$$\Rightarrow y=7 $$
$$\textbf{Hence the lower limit of the class is $$7$$.}$$
Question 5
1 / -0

A grouped frequency table with class intervals of equal sizes using $$250-270$$ ($$270$$ not included in this interval) as one of the class interval is constructed for the following data :
$$268, 220, 368, 258, 242, 310, 272, 342,$$
$$310, 290, 300, 320, 319, 304, 402, 318,$$
$$406, 292, 354, 278, 210, 240, 330, 316,$$
$$406, 215, 258, 236.$$
The frequency of the class $$310-330$$ is:

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

Solution

Numbers between $$310$$ and $$330$$ are $$310, 310, 320, 319, 318, 316.
\space 330$$ will not be included in this interval. Thus, frequency is $$6$$.
Question 6
1 / -0

To draw a histogram to represent the following frequency distribution:
Class Interval
$$5-10$$
$$10-15$$
$$15-25$$
$$25-45$$
$$45-75$$
Frequency
$$6$$
$$12$$
$$10$$
$$8$$
$$15$$
The adjusted frequency for the class $$25-45$$ is:

A
$$6$$

B
$$5$$

C
$$3$$

D
$$2$$

Solution

The frequency distribution will consist of classes with an interval of $$5$$. The class $$25-45$$ includes four such classes with an interval of $$5$$. Thus the adjusted frequency will be $$\dfrac{8}{4}=2$$
Question 7
1 / -0

A grouped frequency distribution table with classes of equal sizes using $$63-72$$ $$(72$$ included$$)$$ as one of the class is constructed for the following data:
$$30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88,\\
40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96,\\
102, 110, 88, 74, 112, 14, 34, 44$$
The number of classes in the distribution will be:

A
$$9$$

B
$$10$$

C
$$11$$

D
$$12$$

Solution

If one the classes is $$63-72$$ then the range of the class is $$10$$ $$(72$$ included$$)$$
Range of the given data points $$= 112-30 = 82$$

Number of classes $$= \dfrac{82}{10} = 8.2$$
Classes required will be $$9$$ after round off.
Question 8
1 / -0

The width of each of five continuous classes in a frequency distribution is $$5$$ and the lower class-limit of the lowest class is $$10$$. The upper class-limit of the highest class is:

A
$$15$$

B
$$25$$

C
$$35$$

D
$$40$$

Solution

Lower class limit $$= 10$$
Width of each class $$= 5$$
Width till the upper class limit for a frequency distribution having $$5$$ classes $$= 5\times 5=25$$

Hence, the upper class limit of highest class $$= 10 +25 =35$$
Question 9
1 / -0

The class marks of a frequency distribution are given as follows: $$15, 20, 25, ...$$
The class corresponding to the class mark $$20$$ is:

A
$$12.5 , 17.5$$

B
$$17.5 , 22.5$$

C
$$18.5 , 21.5$$

D
$$19.5 , 20.5$$

Solution

Class marks are: $$15, 20, 25, ...$$

Lower limit of class $$20$$ is mean of $$15$$ and $$20 = \cfrac{15+20}{2} = 17.5$$
Upper limit of class $$20$$ is the mean of $$20$$ and $$25= \cfrac{20+25}{2} = 22.5 $$

$$20$$ is the class mark for the class interval $$17.5-22.5$$
Question 10
1 / -0

If the mean of the observations:
$$x, x + 3, x + 5, x + 7, x + 10$$ is $$9$$, the mean of the last three observations is

A
$$10\dfrac{1}{3}$$

B
$$10\dfrac{2}{3}$$

C
$$11\dfrac{1}{3}$$

D
$$11\dfrac{2}{3}$$

Solution

Mean of $$x, x + 3, x + 5, x + 7, x + 10 = 9$$
Sum of the terms $$= 5x + 25$$
Mean = $$\dfrac{5x+25}{5} =9$$
$$x +5 =9$$
$$x =4$$
Last three observations are $$9,11,14$$
Mean $$= \dfrac{9+11+14}{3}= 11\dfrac{1}{3}$$

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Re-Attempt Weekly Quiz Competition

Class Interval	$$5-10$$	$$10-15$$	$$15-25$$	$$25-45$$	$$45-75$$
Frequency	$$6$$	$$12$$	$$10$$	$$8$$	$$15$$

Statistics Test - 29

Result

Statistics Test - 29

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Question 1

$$10, 2$$

$$11, 2$$

$$12, 4$$

$$13, 5$$

Solution

Question 2

$$10-20$$

$$20-30$$

Both the intervals

None of these intervals

Solution

Question 3

$$2m + l$$

$$2m - l$$

$$m + l$$

$$m + 2l$$

Solution

Question 4

$$6$$

$$7$$

$$8$$

$$12$$

Solution

Question 5

$$4$$

$$5$$

$$6$$

$$7$$

Solution

Question 6

$$6$$

$$5$$

$$3$$

$$2$$

Solution

Question 7

$$9$$

$$10$$

$$11$$

$$12$$

Solution

Question 8

$$15$$

$$25$$

$$35$$

$$40$$

Solution

Question 9

$$12.5 , 17.5$$

$$17.5 , 22.5$$

$$18.5 , 21.5$$

$$19.5 , 20.5$$

Solution

Question 10

$$10\dfrac{1}{3}$$

$$10\dfrac{2}{3}$$

$$11\dfrac{1}{3}$$

$$11\dfrac{2}{3}$$

Solution

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