Statistics Test

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Statistics Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

Mean of $$50$$ observations was found to be $$80.4$$. But later on, it was discovered that $$96$$ was misread as$$ 69$$ at one place. Find the correct mean.

A
$$75.9$$

B
$$8.4$$

C
$$79$$

D
$$80.94$$

Solution

$$\dfrac { { x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ 50 } }{ 50 } =80.4$$
Say $${ x }_{ 10 }$$ (incorrect value) $$=69$$
$${ x' }_{ 10 }$$ (correct value) $$=96={ x }_{ 10 }+27$$
$${ x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ 50 }=4020$$
Add $$27$$ on both sides
$${ x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ 50 }+27=4020+27=4047$$
$${ x }_{ 1 }+{ x }_{ 2 }+...+{ x' }_{ 10 }+...+{ x }_{ 50 }=4047$$
Actual mean $$=\dfrac { { x }_{ 1 }+{ x }_{ 2 }+...+{ x' }_{ 10 }+...+{ x }_{ 50 } }{ 50 } $$
Therefore, $$\dfrac { 4047 }{ 50 } =80.94$$
Question 2
1 / -0

If the mode of the following data $$10, 11, 12, 10, 15, 14, 15, 13, 12, x, 9, 7$$ is $$15$$, then the value of $$x$$ is:

A
$$10$$

B
$$15$$

C
$$12$$

D
$$\cfrac{21}{2}$$

Solution

The mode is the value which occurs the most often in a given set of data.
In the data set provided here, the mode is 15, hence 15 should occur the most number of times.
$$10, 11, 12, 10, 15, 14, 15, 13, 12, x, 9, 7$$
Excluding $$x$$ and arranging these numbers in ascending order,
$$7, 9, 10, 10, 11, 12, 12, 13, 14, 15, 15$$
Here 10, 12 and 15 are occurring twice.
Since 15 is the mode, it should occur more than twice.
Therefore, the value of $$x$$ should be 15 if the value of mode is 15.
Question 3
1 / -0

The mode of 9, 8, 7, 7, 6, 3, 7, 2, 1, 7, 9 is

A
6

B
7

C
3

D
9

Solution
Question 4
1 / -0

The distribution below gives the weights of $$30$$ students of a class. Find the median weight of the students
Weight (in kg) 40-45
45-50
50-55
55-60
No. of students
2
3
8
6
Weight (in Kg)
60-65
65-70
70-75

No. of students
6
3
2

A
$$46.67$$ g

B
$$12.12$$ kg

C
$$75.12$$ kg

D
$$56.67$$ kg
Question 5
1 / -0

Formula for grouped data of median is

A
$$\dfrac{\dfrac{N}{2}-f}{f}\times C$$

B
$$l +\dfrac{\dfrac{N}{2}-{c}_{f}}{f}\times C$$

C
$$l \times \dfrac{\dfrac{N}{2}-f}{f}$$

D
none of these

Solution

Formula for group data of median $$=l+\dfrac{\dfrac{N}{2}-c_f}{f}\times C$$
Where,
$$l$$ is the lower class boundary of the group containing the median
$$N$$ is the total number of values
$$c_f$$ is the cumulative frequency of the groups before the median group
$$f$$ is the frequency of the median group
$$C$$ is the group width
Question 6
1 / -0

Find mode when median is $$125.6$$ and mean is $$128$$.

A
$$128$$

B
$$126.9$$

C
$$125.6$$

D
$$120.8$$
Question 7
1 / -0

If the values of mode and mean are $$60$$ and $$66$$ respectively, then the value of median is

A
$$12$$

B
$$32$$

C
$$55$$

D
$$64$$

Solution

Since,
$$\text{Mode = 3 Median - 2 Mean}$$
$$\therefore \text{ Median} = \displaystyle \frac{1}{3} (\text{mode} +2 \text{ mean}) = \frac{1}{3} (60 + 2 \times 66) = 64$$

Question 8

1 / -0

The height of $$30$$ boys of a class are given in the following table :

Height in cm	Frequency
120 - 129 130 - 139 140 - 149 150 - 159 150 - 159	2 8 10 7 3

If by joining a boy of height $$140$$ cm, the median of the heights is changed from $$M_1$$ to $$M_2$$, then $$M_1 - M_2$$, in cm is

$$0.1$$

$$- 0.1$$

$$0$$

$$0.2$$

Solution

Height in $$cm$$	Frequency	Cumulative frequency($$c.f.$$)	Actual class limit
$$120-129$$	$$2$$	$$2$$	$$119.5-129.5$$
$$130-139$$	$$8$$	$$10$$	$$129.5-139.5$$
$$140-149$$	$$10$$	$$20$$	$$139.5-149.5$$
$$150-159$$	$$7$$	$$27$$	$$149.5-159.5$$
$$160-169$$	$$3$$	$$30$$	$$159.5-169.5$$

Here $$n = 30$$
$$\therefore \dfrac{n}{2}+ 1 = 15 + 1 = 16$$
$$\therefore$$ $$16$$ is under cumulative frequency $$20$$. So, the median class will be $$140-149$$.
We get, $$L_1 = 139.5, L_2 = 149.5, f = 10, n = 30, c = 10$$

Median $$(M_1) = L_1 + \dfrac{L_2 - L_1}{f}\left ( \dfrac{n}{2} - c\right )$$
$$= 139.5 + \dfrac{10}{10}(15 -10)$$
$$= 139.5 + \dfrac{10}{10} \times 5 = 144.5$$
If we join a boy of height $$140$$ cms, then
$$n = 31, f = 11.$$
$$\therefore$$ Median $$ M_2=139.5 + \dfrac{149.5 - 139.5}{11} (15.5 - 10)$$
$$= 139.5 + \dfrac{10}{11} \times 5.5$$
$$= 144.5$$ cms
Then $$M_1 - M_2 = 144.5 - 144.5 = 0$$

Question 9
1 / -0

Formula for calculating median of the grouped data is

A
$$l - \frac {(\frac {N}{2} - F)}{f} \times C$$

B
$$l + \frac {(\frac {N}{2} - F)}{f} \times C$$

C
$$l + \frac {(\frac {N}{2} + F)}{f} \times C$$

D
non of these
Question 10
1 / -0

If for a moderately skewed distribution, Mode $$= 60$$ and Mean $$=66$$, then median $$=$$

A
$$60$$

B
$$64$$

C
$$68$$

D
None of these

Solution

We have mean $$=66$$, mode $$=60$$
Thus using the relation: mode $$= 3\times$$median $$-$$ $$2\times$$mean
$$\Rightarrow 60 = 3\times median - (2\times66)$$
$$\Rightarrow 60 + (2\times66) = 3\times median$$
$$\Rightarrow$$ $$median$$ $$= \cfrac{60+(2\times 66)}{3}=64$$

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Re-Attempt Weekly Quiz Competition

Weight (in kg)	40-45	45-50	50-55	55-60
No. of students	2	3	8	6
Weight (in Kg)	60-65	65-70	70-75
No. of students	6	3	2

Statistics Test - 26

Result

Statistics Test - 26

Score

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Rank

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Question 1

$$75.9$$

$$8.4$$

$$79$$

$$80.94$$

Solution

Question 2

$$10$$

$$15$$

$$12$$

$$\cfrac{21}{2}$$

Solution

Question 3

6

7

3

9

Solution

Question 4

$$46.67$$ g

$$12.12$$ kg

$$75.12$$ kg

$$56.67$$ kg

Question 5

$$\dfrac{\dfrac{N}{2}-f}{f}\times C$$

$$l +\dfrac{\dfrac{N}{2}-{c}_{f}}{f}\times C$$

$$l \times \dfrac{\dfrac{N}{2}-f}{f}$$

none of these

Solution

Question 6

$$128$$

$$126.9$$

$$125.6$$

$$120.8$$

Question 7

$$12$$

$$32$$

$$55$$

$$64$$

Solution

Question 8

$$0.1$$

$$- 0.1$$

$$0$$

$$0.2$$

Solution

Question 9

$$l - \frac {(\frac {N}{2} - F)}{f} \times C$$

$$l + \frac {(\frac {N}{2} - F)}{f} \times C$$

$$l + \frac {(\frac {N}{2} + F)}{f} \times C$$

non of these

Question 10

$$60$$

$$64$$

$$68$$

None of these

Solution

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