Statistics Test

Result

Statistics Test - 17

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The median of the following discrete series is
$$x_i$$ $$3$$ $$6$$ $$5$$ $$8$$ $$12$$ $$7$$
$$f_i$$ $$5$$ $$2$$ $$4$$ $$6$$ $$7$$ $$6$$

A
$$7$$

B
$$8$$

C
$$9$$

D
$$6$$

Solution

Computation of Median
$$x_i$$ $$f_i$$ $$c.f.$$
$$3$$ $$5$$ $$5$$
$$5$$ $$4$$ $$9$$
$$6$$ $$2$$ $$11$$
$$7$$ $$6$$ $$17$$
$$8$$ $$6$$ $$23$$
$$12$$ $$7$$ $$30$$

Total $$N = 30$$

We have, $$N = 30$$
$$\therefore \quad \displaystyle\frac{N}{2} = 15$$
Cumulative frequency just greater than $$\cfrac N2$$ is $$17$$.
The corresponding value of $$x = 7$$
Hence, median $$= 7$$
Question 2
1 / -0

Median of the observatios $$5$$, $$7$$, $$14$$, $$12$$, $$15$$, $$17$$, $$17$$, $$5$$, $$14$$, $$7$$, $$5$$ is

A
$$14$$

B
$$12$$

C
$$11.8$$

D
$$7$$

Solution

By arranging the given no's in increasing order, we get:
$$ 5,5,5,7,7,12,14,14,15,17,17 $$
Total terms $$= 11$$
since, the middle most term is $$ 6^{th} $$.
$$\therefore $$ value of median is $$12$$
$$\therefore $$ Option $$B$$ is correct.
Question 3
1 / -0

The relationship between mean, median and mode for a moderately skewed distribution is

A
Mode =2 Median - 3 Mean

B
Mode = Median + 2 Mean

C
Mode =3 Median - Mean

D
Mode =3 Median - 2 Mean

Solution

It is well known fact that for moderately skewed distribution,
Mode $$=3\times$$ Median $$-2\times $$ Mean
Question 4
1 / -0

Which of the following gives a correct relation between mean, mode and median?

A
$$Mode = 3Median + 2Mean$$

B
$$Median = Mode +\frac {3}{2} Mean$$

C
$$Mean =\frac {3}{2} Median- \frac {1}{2}Mode $$

D
$$Mode = 2mean - 3 [mean - median]$$

Solution

We know that $$mode=3median-2mean$$

$$\implies median=\dfrac{mode-2mean}3$$

$$\implies median=\dfrac{1}3mode-\dfrac{2}3mean$$

$$\implies mean=\dfrac{3median-mode}2$$

$$\implies mean=\dfrac{3}2median-\dfrac{1}2mode$$
Question 5
1 / -0

Find the mode of $$14$$, $$25$$, $$14$$, $$28$$, $$18$$, $$17$$, $$18$$, $$14$$, $$23$$, $$22$$, $$14$$, $$18$$.

A
$$14$$

B
$$25$$

C
$$28$$

D
$$18$$

Solution

Here, we arrange the data in the ascending order :- $$14,14,14,14,17,18,18,18,22,23,25,28$$
The value $$14$$ has maximum frequency.
Therefore, the mode of the data is $$14$$.
Question 6
1 / -0

Mode of a set of observations is the value which

A
occurs most frequently

B
divides the observations into two equal parts

C
is the mean of the middle two observations

D
is the sum of the observations

Solution

The value which occurs most number of times i.e. most frequently is called the mode of the series.
Question 7
1 / -0

A cricket player scored the following runs in $$11$$ one-day international matches:
$$65$$, $$30$$, $$7$$, $$60$$, $$65$$, $$65$$, $$30$$, $$28$$, $$30$$, $$15$$, $$30$$ Find the modal runs.

A
$$20$$ runs

B
$$25$$ runs

C
$$30$$ runs

D
$$35$$ runs

Solution

From the given observations:
$$65$$, $$30$$, $$7$$, $$60$$, $$65$$, $$65$$, $$30$$, $$28$$, $$30$$, $$15$$, $$30$$
Here, $$30$$ is the most frequent observation. hence, $$30$$ is the mode of the given data.
Question 8
1 / -0

Which of the following is true?

A
Mode = 2Median - Mean

B
Mode = 3Median + 2Mean

C
Mode = 3Median - Mean

D
None of these

Solution

Mode, median and mean of a data series can be represented as:
$$Mode = 3 Median - 2 Mean$$
Question 9
1 / -0

Mode is

A
Most frequent value

B
Least frequent value

C
Middle most value

D
None of these

Solution

The value which occurs the maximum number of times in the dataset or which has the highest frequency is the mode of the dataset.
Question 10
1 / -0

Mode of the following frequency distribution
x: 4 5 6 7 8
f: 6 7 10 8 3

A
$$5$$

B
$$6$$

C
$$8$$

D
$$10$$

Solution

We know mode of any frequency distribution is the term having largest frequency.
Clearly in the given table maximum frequency is $$ 10$$ and corresponding observation is $$6$$. Hence required mode is $$6$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

$$x_i$$	$$3$$	$$6$$	$$5$$	$$8$$	$$12$$	$$7$$
$$f_i$$	$$5$$	$$2$$	$$4$$	$$6$$	$$7$$	$$6$$

$$x_i$$	$$f_i$$	$$c.f.$$
$$3$$	$$5$$	$$5$$
$$5$$	$$4$$	$$9$$
$$6$$	$$2$$	$$11$$
$$7$$	$$6$$	$$17$$
$$8$$	$$6$$	$$23$$
$$12$$	$$7$$	$$30$$
	Total	$$N = 30$$

Statistics Test - 17

Result

Statistics Test - 17

Score

-

Rank

-

SHARING IS CARING

Question 1

$$7$$

$$8$$

$$9$$

$$6$$

Solution

Question 2

$$14$$

$$12$$

$$11.8$$

$$7$$

Solution

Question 3

Mode =2 Median - 3 Mean

Mode = Median + 2 Mean

Mode =3 Median - Mean

Mode =3 Median - 2 Mean

Solution

Question 4

$$Mode = 3Median + 2Mean$$

$$Median = Mode +\frac {3}{2} Mean$$

$$Mean =\frac {3}{2} Median- \frac {1}{2}Mode $$

$$Mode = 2mean - 3 [mean - median]$$

Solution

Question 5

$$14$$

$$25$$

$$28$$

$$18$$

Solution

Question 6

occurs most frequently

divides the observations into two equal parts

is the mean of the middle two observations

is the sum of the observations

Solution

Question 7

$$20$$ runs

$$25$$ runs

$$30$$ runs

$$35$$ runs

Solution

Question 8

Mode = 2Median - Mean

Mode = 3Median + 2Mean

Mode = 3Median - Mean

None of these

Solution

Question 9

Most frequent value

Least frequent value

Middle most value

None of these

Solution

Question 10

$$5$$

$$6$$

$$8$$

$$10$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.