Statistics Test

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Statistics Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

Only __________ can be used for all algebraic calculations.

A
mean

B
mode

C
median

D
All of the above

Solution

Arithmetic mean refers to the average amount in a given group of data. It is defined as the summation of all the observation in the data which is divided by the number of observations in the data. Since mean includes all the values of the series of observation therefore only this average value is included in the algebraic calculations.
Question 2
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Number of Wicket taken by a team in series of $$10$$ matches are $$2, 3, 4, 5, 0, 1, 3, 3, 4, 3$$ find the mode of the set.

A
$$3$$

B
$$2.8$$

C
$$3.3$$

D
$$2.9$$

Solution

Mode is the highest occurring figure in a series. It is the value in a series of observation that repeats maximum number of times and which represents the whole series as most of the values in the series revolves around this value.
Since, 3 is occurring the highest number of times. Therefore, 3 is the mode.
Question 3
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What is the mode value for the numbers $$5, 8, 6, 4, 10, 15, 18, 10$$?

A
$$18$$

B
$$10$$

C
$$14$$

D
None of above

Solution

Mode is the highest occurring figure in a series. It is the value in a series of observation that repeats maximum number of times and which represents the whole series as most of the values in the series revolves around this value. Since in the given series, 10 is occurring the highest number of times. Therefore, 10 is the mode of the series of given observations.
Question 4
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If the frequency of observations $$X_{i}$$ is $$f_{i} (i = 1, 2, ..... n)$$, then which one is correct

A
$$\overline {X} = \dfrac {X_{1}\times X_{2} \times X_{3}\times ......X_{n}}{N}$$

B
$$\overline {X} = \dfrac {X_{1} + X_{2} + X_{3} + ......X_{n}}{N}$$

C
$$\overline {X} = \dfrac {\sum X_{i}}{N}$$

D
$$\overline {X} = \dfrac {\sum f_{i}X_{i}}{\sum f}$$

Solution

Answer is (D) option as it is the formula for calculating mean.
Question 5
1 / -0

The median of the following data is $$525$$. Find the values of $$x$$ and $$y$$, if the total frequency is $$100 $$
Class interval Frequency
$$0-100$$ $$2$$
$$100-200$$ $$5$$
$$200-300$$ $$x$$
$$300-400$$ $$12$$
$$400-500$$ $$17$$
$$500-600$$ $$20$$
$$600-700$$ $$y$$
$$700-800$$ $$9$$
$$800-900$$ $$7$$
$$900-1000$$ $$4$$

A
$$x=15$$ $$y=9$$

B
$$x=5$$ $$y=9$$

C
$$x=15$$ $$y=18$$

D
$$x=9$$ $$y=15$$

Solution

Computation of Median
Class interval Frequency (f) Cumulative frequency (cf)
0-100 2 2
100-200 5 7
200-300 x 7+x
300-400 12 19+x
400-500 17 36+x
500-600 20 56+x
600-700 y 56+x+y
700-800 9 65+x+y
800-900 7 72+x + y
900-1000 4 76+x + y
Total = 100
We have,
$$N= \sum f_i= 100 $$
$$\Rightarrow 76 + x + y = 100 \Rightarrow x+ y = 24$$
It is given that the median is $$525$$. Clearly, it lies in the class $$500-600$$
$$\therefore l= 500, h= 100, f= 20, F= 36+ x $$ and $$ N= 100$$
Now,
$$Median = i+ \frac{\frac{N}{2}-F}{f}\times h$$
$$\Rightarrow 525 = 500 +\frac{50-(36+x)}{20}\times 100$$
$$\Rightarrow 525 -500= (14- x)\times 5$$
$$\Rightarrow 25= 70-5x \Rightarrow 5x = 45 \Rightarrow x=9$$
Putting $$x=9 $$ in$$ x+y= 24$$, we get $$y= 15.$$
Hence, $$x= 9 $$and $$y= 15.$$
Question 6
1 / -0

Find the value of $$p$$ if the Mean of the following distribution is $$50$$
$$x$$ $$10$$ $$30$$ $$50$$ $$70$$ $$90$$
$$y$$ $$17$$ $$5p+3$$ $$32$$ $$7p-11$$ $$19$$

A
$$12$$

B
$$\dfrac{-25}{2}$$

C
$$14$$

D
$$8$$

Solution

$$\\ mean (\frac{10\times17+30(5p+3)+50\times32+70(7p-1)+90\times19}{17+5p+3+32+7p-11+19})\\50=(\frac{170+0150p+90+1600+490p-70+1710}{60+12p})\\50\times(60+12p)=3500+640p\\3000+600p=3500+640p\\\therefore 40p=-500\\\therefore p=(\frac{-25}{2})$$
Question 7
1 / -0

Which of the following has the same mean, median and mode?

A
$$6$$, $$2$$, $$5$$, $$4$$, $$3$$, $$4$$, $$1$$

B
$$4$$,$$2$$,$$2$$, $$1$$,$$3$$,$$2$$,$$3$$

C
$$2$$,$$3$$,$$7$$,$$3$$.$$8$$,$$3$$,$$2$$

D
$$4$$,$$3$$,$$4$$,$$3$$.$$4$$.$$6$$,$$4$$

Solution

Option $$A:6,2,5,4,3,4,1$$
Arranging the numbers in ascending order we get,
$$1,2,3,4,4,5,6$$
$$\Rightarrow$$ $$Mean=\dfrac{1+2+3+4+4+5+6}{7}=\dfrac{25}{7}=3.57$$
$$\Rightarrow$$ The mode of a set of data values is the value that appears most often
$$\Rightarrow$$  Here, number $$4$$ occurs $$2$$ times, so mode is $$4$$
$$\Rightarrow$$  We know, median is a middle number.
$$\Rightarrow$$  Here, $$4th$$ number is middle number which is $$4$$.
$$\therefore$$ Median is $$4$$.
$$\therefore$$ Option $$A$$ does not have same mean, median and mode.

Option $$B:4,2,2,1,3,2,3$$
Arranging the numbers in ascending order we get,
$$1,2,2,2,3,3,4$$
$$\Rightarrow$$ $$Mean=\dfrac{1+2+2+2+3+3+4}{7}=\dfrac{17}{7}=2.42$$
$$\Rightarrow$$  The mode of a set of data values is the value that appears most often
$$\Rightarrow$$  Here, number $$2$$ occurs $$3$$ times, so mode is $$2$$
$$\Rightarrow$$  We know, median is a middle number.
$$\Rightarrow$$  Here, $$4th$$ number is middle number which is $$2$$.
$$\therefore$$ Median is $$2$$.
$$\therefore$$ Option $$B$$ does not have same mean, median and mode.

Option $$C:2,3,7,3,8,3,2$$
Arranging the numbers in ascending order we get,
$$2,2,3,3,3,7,8$$
$$\Rightarrow$$ $$Mean=\dfrac{2+2+3+3+3+7+8}{7}=\dfrac{28}{7}=4$$
$$\Rightarrow$$  The mode of a set of data values is the value that appears most often
$$\Rightarrow$$  Here, number $$3$$ occurs $$3$$ times, so mode is $$3$$
$$\Rightarrow$$  We know, median is a middle number.
$$\Rightarrow$$  Here, $$4th$$ number is middle number which is $$3$$.
$$\therefore$$ Median is $$3$$.
$$\therefore$$ Option $$C$$ does not have same mean, median and mode.

Option $$D:4,3,4,3,4,6,4$$
Arranging the numbers in ascending order we get,
$$3,3,4,4,4,4,6$$
$$\Rightarrow$$ $$Mean=\dfrac{3+3+4+4+4+4+6}{7}=\dfrac{28}{7}=4$$
$$\Rightarrow$$  The mode of a set of data values is the value that appears most often
$$\Rightarrow$$  Here, number $$4$$ occurs $$4$$ times, so mode is $$4$$
$$\Rightarrow$$  We know, median is a middle number.
$$\Rightarrow$$  Here, $$4th$$ number is middle number which is $$4$$.
$$\therefore$$ Median is $$4$$.
$$\therefore$$ Option $$D$$ have same mean, median and mode.
Question 8
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If the mean of the following data is $$20.2$$ then the value of $$p$$ is
$$x$$ $$10$$ $$15$$ $$20$$ $$25$$ $$30$$
$$f$$ $$6$$ $$8$$ $$p$$ $$10$$ $$6$$

A
$$40$$

B
$$20$$

C
$$60$$

D
None

Solution

Given:
Mean, $$\overline x = 20.2$$

$$x_i$$ $$f_i$$ $$f_ix_i$$
$$10$$ $$6$$ $$60$$
$$15$$ $$8$$ $$120$$
$$20$$ $$p$$ $$20p$$
$$25$$ $$10$$ $$250$$
$$30$$ $$6$$ $$180$$
$$\sum f_i=30+p$$ $$\sum f_ix_i=610+20p$$
As we know that,
$$\overline x = \dfrac{\sum f_ix_i}{\sum f_i}$$
$$ \Rightarrow 20.2 = \dfrac{{610 + 20p}}{{30 + p}}$$
$$ \Rightarrow 606 + 20.2p = 610 + 20p$$
$$ \Rightarrow 20.2p - 20p = 610 - 606$$
$$ \Rightarrow 0.2p = 4$$
$$ \Rightarrow p = 20$$

Hence, option $$B$$ is correct.
Question 9
1 / -0

If the difference between mean and mode is $$63$$, the difference between mean and median is?

A
$$189$$

B
$$21$$

C
$$31.5$$

D
$$48.5$$

Solution

$$\begin{array}{l} \left( { mean-mode } \right) =3\left( { mean-median } \right) \\ 63=3\left( { mean-median } \right) \\ 21=mean-median \end{array}$$
Question 10
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The mean of the following data is $$50.$$ Find the value of a and hence the frequencies of $$30$$ and $$70.$$
$$\begin{array}{*{20}{c}}X&{10}&{30}&{50}&{70}&{90}\\F&{17}&{5a + 3}&{32}&{7a -11}&{19}\end{array}$$

A
$$28$$ and $$34$$

B
$$68$$ and $$24$$

C
$$28$$ and $$24$$

D
None of these

Solution

X $$10$$ $$30$$ $$50$$ $$70$$ $$90$$
F $$17$$ $$5 \alpha + 3$$ $$32$$ $$7\alpha - 11$$ $$19$$
Mean = $$\dfrac{\displaystyle\sum fixi}{\displaystyle\sum fi} = 50$$
$$50 = \dfrac{10 \times 17 + 30 (5 \alpha + 3) + 50 \times 32 + 70 (7\alpha -11) + 90 \times 19}{17 + 5 \alpha + 3 + 32 + 7 \alpha - 11 + 19}$$
$$50 (12 \alpha + 60) = 640 \alpha + 2800$$
$$3000 + 600 \alpha = 640 \alpha + 2800$$
$$40 \alpha = 200$$
$$\alpha = 5$$
frequency of $$30 = 5 \alpha + 3 = 25 + 3 = 28$$
frequency of $$70 = 7 \alpha - 11 = 35 - 11 = 24$$.

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Class interval	Frequency
$$0-100$$	$$2$$
$$100-200$$	$$5$$
$$200-300$$	$$x$$
$$300-400$$	$$12$$
$$400-500$$	$$17$$
$$500-600$$	$$20$$
$$600-700$$	$$y$$
$$700-800$$	$$9$$
$$800-900$$	$$7$$
$$900-1000$$	$$4$$

Class interval	Frequency (f)	Cumulative frequency (cf)
0-100	2	2
100-200	5	7
200-300	x	7+x
300-400	12	19+x
400-500	17	36+x
500-600	20	56+x
600-700	y	56+x+y
700-800	9	65+x+y
800-900	7	72+x + y
900-1000	4	76+x + y
		Total = 100

$$x$$	$$10$$	$$30$$	$$50$$	$$70$$	$$90$$
$$y$$	$$17$$	$$5p+3$$	$$32$$	$$7p-11$$	$$19$$

$$x$$	$$10$$	$$15$$	$$20$$	$$25$$	$$30$$
$$f$$	$$6$$	$$8$$	$$p$$	$$10$$	$$6$$

$$x_i$$	$$f_i$$	$$f_ix_i$$
$$10$$	$$6$$	$$60$$
$$15$$	$$8$$	$$120$$
$$20$$	$$p$$	$$20p$$
$$25$$	$$10$$	$$250$$
$$30$$	$$6$$	$$180$$
	$$\sum f_i=30+p$$	$$\sum f_ix_i=610+20p$$

Statistics Test - 53

Result

Statistics Test - 53

Score

-

Rank

-

SHARING IS CARING

Question 1

mean

mode

median

All of the above

Solution

Question 2

$$3$$

$$2.8$$

$$3.3$$

$$2.9$$

Solution

Question 3

$$18$$

$$10$$

$$14$$

None of above

Solution

Question 4

$$\overline {X} = \dfrac {X_{1}\times X_{2} \times X_{3}\times ......X_{n}}{N}$$

$$\overline {X} = \dfrac {X_{1} + X_{2} + X_{3} + ......X_{n}}{N}$$

$$\overline {X} = \dfrac {\sum X_{i}}{N}$$

$$\overline {X} = \dfrac {\sum f_{i}X_{i}}{\sum f}$$

Solution

Question 5

$$x=15$$ $$y=9$$

$$x=5$$ $$y=9$$

$$x=15$$ $$y=18$$

$$x=9$$ $$y=15$$

Solution

Question 6

$$12$$

$$\dfrac{-25}{2}$$

$$14$$

$$8$$

Solution

Question 7

$$6$$, $$2$$, $$5$$, $$4$$, $$3$$, $$4$$, $$1$$

$$4$$,$$2$$,$$2$$, $$1$$,$$3$$,$$2$$,$$3$$

$$2$$,$$3$$,$$7$$,$$3$$.$$8$$,$$3$$,$$2$$

$$4$$,$$3$$,$$4$$,$$3$$.$$4$$.$$6$$,$$4$$

Solution

Question 8

$$40$$

$$20$$

$$60$$

None

Solution

Question 9

$$189$$

$$21$$

$$31.5$$

$$48.5$$

Solution

Question 10

$$28$$ and $$34$$

$$68$$ and $$24$$

$$28$$ and $$24$$

None of these

Solution

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