Statistics Test

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Statistics Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

Given the following distribution table:
Class
$$0-5$$
$$5-10$$
$$10-15$$
$$15-20$$
$$20-25$$
Frequency
$$10$$
$$15$$
$$12$$
$$20$$
$$9$$
The sum of lower limits of the Median class and the Modal class is:

A
$$15$$

B
$$25$$

C
$$30$$

D
$$35$$

Solution

Answer:-
The cumulative frequency Table is:-

Class Frequency Cumulative frequency
0-5 10 10
5-10 15 25
10-15 12 37
15-20 20 57
20-25 9 66
Here, $$ \cfrac{N}{2} = \cfrac{66}{2} = 33 $$
Thus $$37$$ is just greater than $$33$$,
$$\therefore$$ The median class is "$$10-15.$$"
The lower limit of the median class is 10.
Modal class = class with maximum frequency.
$$ \therefore $$ Modal class is $$15-20.$$
The lower limit of the modal class is $$15$$.
Sum of lower limits of median class and modal class = $$10 + 15 = 25$$

Question 2

1 / -0

If the median of the distribution is $$28.5$$, find thevalues of $$x_1$$ and $$x_2$$

Class interval	Frequency
0-10	5
10-20	$$x_1$$
20-30	20
30-40	15
40-50	$$x_2$$
50-60	5
	Total $$=$$ 60

$$8, 7$$

$$9, 7$$

$$6, 8$$

$$7, 9$$

Solution

Answer:-

Class Interval	Frequency (F)	Cumulative frequency (C. F.)
0-10	5	5
10-20	$$ x_1 $$	5 + $$x_1$$
20-30	20	25 + $$x_1$$
30-40	15	40 + $$x_1$$
40-50	$$ x_2 $$	40 + $$ x_1 + x_2 $$
50-60	5	45 + $$ x_1 + x_2 $$
Total	60

Given:- $$ \Sigma F = 60 $$

$$ \therefore x_1 + x_2 + 45 = 60 $$

$$ \Rightarrow x_1 + x_2 = 15 \longrightarrow eq. (i) $$

Given: - Median = $$28.5$$

Using media:- $$ L + \cfrac{\cfrac{N}{2} - C}{f} \times w $$

Where $$L$$ is lower limit of class $$20-30$$

$$f$$ is frequency of class $$20-30,$$

$$w$$ is width of class

$$C$$ is cumulative frequency of previous class and$$ N$$ = $$\Sigma F$$

$$ \Rightarrow $$ $$L= 20$$, $$ \cfrac{N}{2} = \cfrac{60}{2} $$ = $$30$$, $$C = x_1 + 5 ,$$ $$w = 30 - 20 = 10$$, $$f = 20$$

$$ \Rightarrow M = 20 + \cfrac{30 - (x_1 + 5)}{20} \times 10 $$

$$ 28.5 = 20 + \cfrac{25 - x_1}{20} \times 10 $$

$$ 8.5 = \cfrac{25 - x_1}{2} $$

$$ x_1 = 8$$

From eq. (i), we get

$$ x_1 + x_2 = 15 $$

$$ 8 + x_2 = 15 $$

$$ x_2 = 7$$

$$\therefore $$ value of $$ x_1 \& x_2 $$ are $$8$$ and $$7$$ respectively.

Question 3
1 / -0

If $$x_i$$'s are the mid points of the class intervals of grouped data, $$f_1$$'s are the corresponding frequencies and $$\bar x$$ is the mean, then $$\sum (f_i x_i - \bar x)$$ is equal to

A
$$0$$

B
$$-1$$

C
$$1$$

D
$$2$$

Solution

Given that, $$x_i$$ are the mid points of the class intervals.
$$f_i$$ are the corresponding frequencies.
The mean $$\bar x$$ for the grouped data is $$\bar x=\dfrac{\sum f_ix_i}{n}$$

$$\implies \bar xn=\sum f_ix_i$$ ------(1)

$$\sum (f_ix_i-\bar x)=\sum f_ix_i-\sum \bar x$$

$$\implies \sum (f_ix_i-\bar x)=\bar xn-\sum \bar x$$ (from (1))

$$\implies \sum (f_ix_i-\bar x)=\bar xn-\bar xn=0$$
Question 4
1 / -0

If the sum of the mode and mean of the certain frequency distribution in $$129$$ and the median of the observations is $$63$$, mode and mean are respectively

A
$$69$$ and $$60$$

B
$$65$$ and $$64$$

C
$$68$$ and $$61$$

D
none of these

Solution

Given, Mode$$+$$ Mean $$=$$ 129 .....(1)
and Median $$=63$$,
Also we know, Mode $$=$$ 3. Median $$-$$ 2. Mean
$$\Rightarrow$$ Mode$$=3\times 63-2.$$ Mean $$=189-2.$$Mean ..(2)
Solving (1) and (2) we get, Mean$$=60$$ and Mode $$=69$$.
Question 5
1 / -0

Family size $$1-3$$ $$3-5$$ $$5-7$$ $$7-9$$ $$9-11$$
No. of families $$14$$ $$16$$ $$4$$ $$4$$ $$2$$
A survey conducted on $$40$$ households in a locality by a group of students resulted in the following frequency table for the number of family members in a household -
Find the mode of this data

A
$$3.29$$

B
$$4.12$$

C
$$2.56$$

D
$$6.12$$

Solution

Here the maximum class frequency is $$16$$, and the class corresponding to this frequency is $$3-5$$. So, the modal class is $$3-5$$.

Now, in the modal class $$3-5$$, lower limit of the modal class is $$(l)=3$$,

class size $$(h)=$$ 2, frequency of the modal class $$(f_1)=16$$,

frequency of class preceding the modal class $$(f_0)=14$$,

frequency of class succeeding the modal class $$(f_2) = 4$$,

Substituting these values in the formula, we have
$$Mode = \displaystyle l + \left ( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right ) \times h$$

$$= 3 + \displaystyle \left ( \frac{16-14}{2 \times 16 - 14 - 4} \right ) \times 2 \\= 3 + \left ( \dfrac{2}{14} \right ) \times 2 \\= 3 + \dfrac{4}{14} \\= 3.29$$
Question 6
1 / -0

If in a moderately skewed distribution the values of mode and mean are $$6\lambda$$ and $$9\lambda$$ respectively, then the value of the median is

A
$$8\lambda$$

B
$$7\lambda$$

C
$$6\lambda$$

D
$$5\lambda$$

Solution

Using, Mode $$=$$ 3. Median $$-$$ 2. Mean
$$\Rightarrow 6\lambda =$$ 3.Median $$-2.9\lambda$$
$$\therefore$$ Median $$= 8 \lambda$$
Question 7
1 / -0

In a distribution the values of mean and median are $$5$$ and $$6$$ respectively. The value of mode in such a situation is approximately equal to

A
$$8$$

B
$$11$$

C
$$16$$

D
none of these

Solution

Relationship between mean, mode and median is given by:
$$mode = 3\times (median) - 2 \times( mean)$$
Given, mean $$=5$$ and median $$ = 6$$
$$\therefore mode = 3\times (median) - 2 \times( mean) $$
$$= 3(6)-2(5)$$
$$=8$$
Question 8
1 / -0

If the difference between the mode and median is $$2$$, then the difference between the median and mean (in the given order) is?

A
$$2$$

B
$$4$$

C
$$1$$

D
$$0$$

Solution

We know that, Mode $$=3\times$$Median $$- 2\times$$Mean
$$\Rightarrow$$ Mode $$-$$ Median $$= 2\times($$Median $$-$$ Mean$$)$$
$$\Rightarrow 2=2\times($$Median $$-$$ Mean$$)$$
$$\therefore$$ Median $$-$$ Mean $$= 1$$

Hence option (C) is correct
Question 9
1 / -0

The median class in the following frequency distribution is:
Class Interval: 0-10 10-20 20-30 30-40 40-50
Frequency 12 13 25 20 10

A
$$10-20$$

B
$$20-30$$

C
$$30-40$$

D
none of these

Solution

Class Frequency Cumulative Frequency
0-10 12 12
10-20 13 25
20-30 25 50
30-40 20 60
40-50 10 70

Now $$\dfrac{N}{2}=35$$. Hence cumulative frequency just greater than $$35$$ is $$50$$. The corresponding class is $$20-30$$.
Question 10
1 / -0

The mode of the following distribution is
Class Interval: 1-5 6-10 11-15 16-20 21-25
Frequency: 4 7 10 8 6

A
$$14.5$$

B
$$16.5$$

C
$$10.5$$

D
$$13.5$$

Solution

Modal group = 11-15 , since it has the highest frequency i.e $$10$$.
Estimated Mode $$M=L+\dfrac{f_m-f_{{m-1}}}{f_m-f_{m-1}+f_m-f_{m+1}}\times w$$

Lower class boundary of modal group $$L=10.5$$
Frequency of group before the modal group $$f_{m-1}=7$$
Frequency of Modal group $$f_m=10$$
Frequency of group after the modal group $$f_{m+1}=8$$
Group width $$w=5$$

$$\therefore M= 10.5 + \dfrac{10-7}{10-7+10-8}\times 5$$
$$=10.5+\dfrac{3}{5}\times5$$
$$=13.5$$
hence, option $$D$$ is correct.

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Class	$$0-5$$	$$5-10$$	$$10-15$$	$$15-20$$	$$20-25$$
Frequency	$$10$$	$$15$$	$$12$$	$$20$$	$$9$$

Class	Frequency	Cumulative frequency
0-5	10	10
5-10	15	25
10-15	12	37
15-20	20	57
20-25	9	66

Family size	$$1-3$$	$$3-5$$	$$5-7$$	$$7-9$$	$$9-11$$
No. of families	$$14$$	$$16$$	$$4$$	$$4$$	$$2$$

Class Interval:	1-5	6-10	11-15	16-20	21-25
Frequency:	4	7	10	8	6

Statistics Test - 27

Result

Statistics Test - 27

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Rank

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SHARING IS CARING

Question 1

$$15$$

$$25$$

$$30$$

$$35$$

Solution

Question 2

$$8, 7$$

$$9, 7$$

$$6, 8$$

$$7, 9$$

Solution

Question 3

$$0$$

$$-1$$

$$1$$

$$2$$

Solution

Question 4

$$69$$ and $$60$$

$$65$$ and $$64$$

$$68$$ and $$61$$

none of these

Solution

Question 5

$$3.29$$

$$4.12$$

$$2.56$$

$$6.12$$

Solution

Question 6

$$8\lambda$$

$$7\lambda$$

$$6\lambda$$

$$5\lambda$$

Solution

Question 7

$$8$$

$$11$$

$$16$$

none of these

Solution

Question 8

$$2$$

$$4$$

$$1$$

$$0$$

Solution

Question 9

$$10-20$$

$$20-30$$

$$30-40$$

none of these

Solution

Question 10

$$14.5$$

$$16.5$$

$$10.5$$

$$13.5$$

Solution

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