Statistics Test

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Statistics Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

Find the median for the following data$$:$$
X
$$6$$
$$12$$
$$18$$
$$24$$
F
$$2$$
$$4$$
$$6$$
$$8$$

A
$$17$$

B
$$18$$

C
$$19$$

D
$$20$$

Solution

X
Frequency(f)
Cumulative frequency(cf)
$$6$$
$$2$$
$$2$$
$$12$$
$$4$$
$$2+4=6$$
$$18$$
$$6$$
$$6+6=12$$
$$24$$
$$8$$
$$12+8=20$$

$$n=\displaystyle\Sigma f=20$$
$$\displaystyle\frac{n}{2}=\frac{10}{2}=10$$

$$\displaystyle\frac{n}{2}=\frac{10}{2}=10$$
Therefore,

the corresponding value of $$10$$ is $$18$$.
Question 2
1 / -0

Find mode for the following data$$:$$
Size
$$2-5$$
$$5-8$$
$$8-11$$
$$11-14$$
Number of shoes
$$1$$
$$4$$
$$8$$
$$7$$

A
$$10.4$$

B
$$11.4$$

C
$$12.4$$

D
$$13.4$$

Solution

Here frequency of class interval $$8-11$$ is maximum.
So, it is the modal class
Now $$ l $$ = lower limit of modal class $$= 8$$
$$f_1 =$$ frequency of modal class $$= 8$$
$$f_0 =$$ frequency of class preceding the modal class $$= 4$$
$$f_2 =$$ frequency of class succeeding the modal class $$= 7$$
$$h =$$ class interval width $$= 3$$
So, Mode $$=l+\left(\displaystyle\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$$
$$=8+\left(\displaystyle\frac{8-4}{2\times 8-4-7}\right)\times 3$$
$$=8+\left(\displaystyle\frac{4}{16-11}\right)\times 3$$
$$=8+2.4$$
$$=10.4$$
Mode$$=10.4$$
Question 3
1 / -0

Find the mode for the following data$$:$$
Students $$10$$ $$14$$ $$20$$ $$30$$ $$60$$
Frequency $$2$$ $$2$$ $$4$$ $$7$$ $$8$$

A
$$10$$

B
$$20$$

C
$$60$$

D
$$30$$

Solution

Mode for the discrete series is the variable which has the highest frequency.
From the given distribution table, the highest frequency is $$8$$ which is associated with the variable $$60$$
Therefore, the mode is $$60$$
Question 4
1 / -0

Calculate the mode for the following data$$:$$
Daily income
$$100-200$$
$$200-300$$
$$300-400$$
$$400-500$$
$$500-600$$
$$600-700$$
Workers
$$12$$
$$23$$
$$45$$
$$50$$
$$30$$
$$12$$

A
$$410$$

B
$$405$$

C
$$420$$

D
$$440$$

Solution

Here frequency of class interval $$400-500$$ is maximum.
So, it is the modal class
Now, $$ l $$ = lower limit of modal class = 400
$$f_1 =$$ frequency of modal class $$= 50$$
$$f_0 =$$ frequency of class preceding the modal class $$= 45$$
$$f_2 = $$frequency of class succeeding the modal class $$= 30$$
$$h =$$ class interval width $$= 100$$
So, Mode $$=l+\left(\displaystyle\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$$
$$=400+\left(\displaystyle\frac{50-45}{2\times 50-45-30}\right)\times 100$$
$$=400+\left(\displaystyle\frac{5}{100-75}\right)\times 100$$
$$=400+20$$
$$=420$$
Mode$$=420$$
Question 5
1 / -0

Find the mode for the following data $$:$$
Class interval
$$0-2$$
$$2-4$$
$$4-6$$
$$6-8$$
$$8-10$$
Frequency
$$1$$
$$3$$
$$6$$
$$5$$
$$2$$

A
$$5.2$$

B
$$5.3$$

C
$$5.4$$

D
$$5.5$$

Solution

Here frequency of class interval $$4-6$$ is maximum.
So, it is the modal class
Now $$ l $$ = lower limit of modal class $$= 4$$
$$f_1 =$$ frequency of modal class $$= 6$$
$$f_0 =$$ frequency of class preceding the modal class $$= 3$$
$$f_2 =$$ frequency of class succeeding the modal class $$= 5$$
$$h =$$ class interval width $$= 2$$
So, Mode $$=l+\left(\displaystyle\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$$
$$=4+\left(\displaystyle\frac{6-3}{2\times 6-3-5}\right)\times 2$$
$$=4+\left(\displaystyle\frac{3}{12-8}\right)\times 2$$
$$=4+1.5$$
$$=5.5$$
Mode$$=5.5$$

Question 6

1 / -0

The time taken by a group of people to run across the street is given below. Find the median.

Time(min)	$$10$$	$$20$$	$$25$$	$$30$$	$$45$$
People	$$1$$	$$2$$	$$5$$	$$6$$	$$7$$

$$10$$ min

$$20$$ min

$$30$$ min

$$40$$ min

Solution

Time(min)	People(f)	Cumulative frequency(cf)
$$10$$	$$1$$	$$1$$
$$20$$	$$2$$	$$1+2=3$$
$$25$$	$$5$$	$$3+5=8$$
$$30$$	$$6$$	$$8+6=14$$
$$45$$	$$7$$	$$14+7=21$$
	$$n=\displaystyle\Sigma f=21$$ $$\displaystyle\frac{n}{2}=\frac{21}{2}=10.5$$

$$\displaystyle\frac{n}{2}=\frac{21}{2}=10.5$$
Therefore, the corresponding value of time $$10.5$$ is $$30$$min.

Question 7
1 / -0

Find the approximate value of mode for the following data $$:$$
Class interval
$$7-14$$
$$14-21$$
$$21-28$$
$$28-35$$
$$35-42$$
Frequency
$$2$$
$$4$$
$$3$$
$$10$$
$$1$$

A
$$30$$

B
$$32$$

C
$$31$$

D
$$35$$

Solution

Here frequency of class interval $$28-35$$ is maximum.
So, it is the modal class
Now $$ l $$ = lower limit of modal class $$= 28$$
$$f_1 =$$ frequency of modal class $$= 10$$
$$f_0 = $$frequency of class preceding the modal class $$= 3$$
$$f_2 =$$ frequency of class succeeding the modal class $$= 1$$
$$h =$$ class interval width $$= 7$$
So, Mode $$=l+\left(\displaystyle\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$$
$$=28+\left(\displaystyle\frac{10-3}{2\times 10-3-1}\right)\times 7$$
$$=28+\left(\displaystyle\frac{7}{20-4}\right)\times 7$$
$$=28+3.0625$$
$$=31.06$$
Mode$$\sim 31$$
Question 8
1 / -0

What is the median class for the following data given below?
Height(cm)
$$0-10$$
$$10-20$$
$$20-30$$
$$30-40$$
$$40-50$$
$$50-60$$
Number of players
$$3$$
$$6$$
$$2$$
$$5$$
$$10$$
$$4$$

A
$$0-10$$

B
$$10-20$$

C
$$20-30$$

D
$$30-40$$

Solution

To find the median class:
First add the total number of frequencies.
Here, $$f = 30$$
So, $$\dfrac f2= \dfrac{(Total \quad number \quad of \quad frequency )}{2}$$
$$=\dfrac{ (30 )}{2}$$
$$ = 15$$
The cumulative frequency to which $$\dfrac{f}{2}$$ belongs, the value $$15$$ lies in the class interval $$30-40$$ from the cummulative frequency table.

So the median class is $$30-40$$
Question 9
1 / -0

Calculate the mode for the following data$$:$$
Score $$14$$ $$16$$ $$18$$ $$20$$
Frequency $$2$$ $$4$$ $$4$$ $$8$$

A
$$14$$

B
$$16$$

C
$$18$$

D
$$20$$

Solution

Mode for the discrete series is the variable which has the highest frequency.
From the given distribution table, the highest frequency is $$8$$ which is associated with the variable $$20$$
Therefore, the mode is $$20$$
Question 10
1 / -0

Find the median number of newspaper purchased in a story by $$9$$ customers.
$$2, 4, 5, 1, 6, 8, 9, 4, 7$$

A
$$2$$

B
$$4$$

C
$$5$$

D
$$7$$

Solution

The median of a set of data is the middlemost number in the set.
So, first arrange the data in order.
$$1, 2, 4, 4, 5, 6, 7, 8, 9$$
The median is $$5$$.

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X	Frequency(f)	Cumulative frequency(cf)
$$6$$	$$2$$	$$2$$
$$12$$	$$4$$	$$2+4=6$$
$$18$$	$$6$$	$$6+6=12$$
$$24$$	$$8$$	$$12+8=20$$
	$$n=\displaystyle\Sigma f=20$$ $$\displaystyle\frac{n}{2}=\frac{10}{2}=10$$

Size	$$2-5$$	$$5-8$$	$$8-11$$	$$11-14$$
Number of shoes	$$1$$	$$4$$	$$8$$	$$7$$

Students	$$10$$	$$14$$	$$20$$	$$30$$	$$60$$
Frequency	$$2$$	$$2$$	$$4$$	$$7$$	$$8$$

Daily income	$$100-200$$	$$200-300$$	$$300-400$$	$$400-500$$	$$500-600$$	$$600-700$$
Workers	$$12$$	$$23$$	$$45$$	$$50$$	$$30$$	$$12$$

Class interval	$$0-2$$	$$2-4$$	$$4-6$$	$$6-8$$	$$8-10$$
Frequency	$$1$$	$$3$$	$$6$$	$$5$$	$$2$$

Class interval	$$7-14$$	$$14-21$$	$$21-28$$	$$28-35$$	$$35-42$$
Frequency	$$2$$	$$4$$	$$3$$	$$10$$	$$1$$

Height(cm)	$$0-10$$	$$10-20$$	$$20-30$$	$$30-40$$	$$40-50$$	$$50-60$$
Number of players	$$3$$	$$6$$	$$2$$	$$5$$	$$10$$	$$4$$

Statistics Test - 20

Result

Statistics Test - 20

Score

-

Rank

-

SHARING IS CARING

Question 1

$$17$$

$$18$$

$$19$$

$$20$$

Solution

Question 2

$$10.4$$

$$11.4$$

$$12.4$$

$$13.4$$

Solution

Question 3

$$10$$

$$20$$

$$60$$

$$30$$

Solution

Question 4

$$410$$

$$405$$

$$420$$

$$440$$

Solution

Question 5

$$5.2$$

$$5.3$$

$$5.4$$

$$5.5$$

Solution

Question 6

$$10$$ min

$$20$$ min

$$30$$ min

$$40$$ min

Solution

Question 7

$$30$$

$$32$$

$$31$$

$$35$$

Solution

Question 8

$$0-10$$

$$10-20$$

$$20-30$$

$$30-40$$

Solution

Question 9

$$14$$

$$16$$

$$18$$

$$20$$

Solution

Question 10

$$2$$

$$4$$

$$5$$

$$7$$

Solution

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