Statistics Test

Result

Statistics Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

Find the mode for the following data$$:$$
Farm size $$12$$ $$13$$ $$14$$ $$15$$ $$17$$ $$18$$ $$19$$ $$20$$
Number of animals $$3$$ $$4$$ $$4$$ $$2$$ $$1$$ $$2$$ $$4$$ $$1$$

A
$$13$$

B
$$14$$

C
$$19$$

D
All the above

Solution

Mode for the discrete series is the variable which has the highest frequency.
From the given distribution table, the highest frequency is $$4$$ which is associated with the variables $$13,14$$ and $$19$$
Therefore, the mode is $$13,14$$ and $$19$$
Question 2
1 / -0

The following is a frequency table of the score obtained in a science quiz competition. Find the median score.
score
$$10$$
$$15$$
$$20$$
$$25$$
$$30$$
frequency
$$2$$
$$3$$
$$5$$
$$6$$
$$4$$

A
$$22$$

B
$$22.5$$

C
$$20$$

D
$$25$$

Solution

For the discrete data series, median is the $$(\dfrac{n+1}{2})^{th}$$ value.

Where $$n$$ is the number of observations.

Therefore, $$n=5$$

Hence, $$median=\dfrac{5+1}2=\dfrac{6}2=3^{rd}$$ value

Therefore, $$median=20$$
Question 3
1 / -0

Calculate mode for the following data shows the number of colour pencils the students have in a class.
Colour Pencils $$0-5$$ $$5-10$$ $$10-15$$ $$15-20$$ $$20-25$$ $$25-30$$
Number of students $$14$$ $$16$$ $$17$$ $$13$$ $$8$$ $$12$$

A
$$11$$

B
$$15$$

C
$$19$$

D
$$16$$

Solution

The lower limit ( l ) of the modal class $$= 10,$$
The class size ( h) $$= 5,$$
The frequency $$(f_1)$$ of modal class $$=17$$,
The frequency $$(f_0)$$ of the class preceding the modal class $$= 16$$,
The frequency $$(f_2)$$ of the class succeeding the modal class $$=13.$$
Now, using the formula Mode$$=l+\left(\displaystyle\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$$
$$=60+\left(\displaystyle\frac{17-16}{(2\times 17)-16-13}\right)\times 5$$
$$=10+0.2\times 5$$
$$=10+1$$
$$=11$$
Therefore, mode of colour pencil students have is $$11$$

Question 4

1 / -0

A survey was conducted to identify vegetarians in a group of people. Find the mode of the data.

Group	$$5-7$$	$$7-9$$	$$9-11$$	$$11-13$$	$$13-15$$	$$15-17$$
Frequency	$$6$$	$$7$$	$$3$$	$$7$$	$$12$$	$$9$$

$$15.25$$

$$12.25$$

$$14.25$$

$$13.25$$

Solution

Group	frequency
5-7	6
7-9	7
9-11	3
11-13	7
13-15	12
15-17	9

Maximum frequency$$=12$$

Modal class$$=13-15$$

$$l=3$$, $$h=12$$

$${ f }_{ I }=12{ ,f }_{ 0 }{ =7,f }_{ 2 }=9$$

Mode $$=\left( \cfrac { { f }_{ I }{ -f_{ 0 } } }{ { 2f }_{ I }-{ f }_{ 0 }{ -f }_{ 2 } } \right) \times h$$

$$=13+\left( \cfrac { 12-7 }{ 24-7-9 } \right) \times 2$$

$$=13+\left( \cfrac { 5 }{ 8 } \right) \times 2$$

$$=13+\left( \cfrac { 5 }{ 4 } \right) $$

$$=13+1.25$$

$$=14.25$$

Question 5

1 / -0

In a school the mark distribution of $$25$$ students in a mathematics examination is given below. Calculate it's mode.

Marks	$$30-40$$	$$40-50$$	$$50-60$$	$$60-70$$	$$70-80$$	$$80-90$$
No. of students	$$3$$	$$2$$	$$4$$	$$10$$	$$4$$	$$2$$

$$65$$

$$32$$

$$34$$

$$31$$

Solution

Marks	Number of Students
$$30-40$$	$$3$$
$$40-50$$	$$2$$
$$50-60$$	$$4$$
$$60-70$$	$$10$$
$$70-80$$	$$4$$
$$80-90$$	$$2$$

$$Maximum\ frequency=10$$

Modal class $$=60-70$$

Lower value(l) $$=60$$

($${ f }_{ 1 }$$) frequency of modal class $$=10$$

($${ f }_{ 0 }$$) frequency of preceeding class $$4$$

($${ f }_{ 2 }$$) frequency of succeeding class $$=4$$

$$h=10$$

$$Mode=l+\left( \cfrac { { f }_{ 1 }-{ f }_{ 0 } }{ { 2f }_{ 1 }-{ f }_{ 0- }{ f }_{ 2 } } \right) \times h\\ \quad =60+\left( \cfrac { 10-4 }{ 20-4-4 } \right) \times 10\\ =60+\cfrac { 6 }{ 12 } \times 10\\ =60+\cfrac { 60 }{ 12 } \\ =60+5\\ =65Ans$$

Question 6

1 / -0

The following data shows distance covered by horses in a tourist spot. Find mode of the distance travelled by the horses.

Distance(km)	$$0-20$$	$$20-40$$	$$40-60$$	$$60-100$$	$$100-120$$
Number of horses	$$20$$	$$40$$	$$30$$	$$50$$	$$40$$

$$69.2$$km

$$70.2$$km

$$72.2$$km

$$73.2$$km

Solution

Distance	Number of horses
$$0-20$$	$$20$$
$$20-40$$	$$40$$
$$40-60$$	$$30$$
$$60-80$$	$$50$$
$$80-100$$	$$40$$

Maximum frequency $$=50$$

$$Modal class =60-100$$

Lower value(l) $$=60$$, $${ f }_{ 1 }=50$$

$${ f }_{ 0- }=30$$, $${ f }_{ 2 }=40$$, $$h=20$$

$$Mode=l+\left( \cfrac { { f }_{ 1 }-{ f }_{ 0 } }{ { 2f }_{ 1 }-{ f }_{ 0- }{ f }_{ 2 } } \right) \times h\\ \quad =60+\left( \cfrac { 50-30 }{ 100-30-40 } \right) \times 20\\ =60+\cfrac { 20 }{ 30 } \times 20\\ =60+\cfrac { 40 }{ 3 } \\ =60+13.2\\ =73.2Ans$$

Question 7

1 / -0

Find the median for the following data:

Number of animals	$$10$$	$$15$$	$$25$$	$$45$$	$$50$$
Average life span (years)	$$2$$	$$1$$	$$3$$	$$4$$	$$5$$

$$10$$

$$15$$

$$25$$

$$45$$

Solution

Number of Animal	Average life span	Position
$$10$$	$$2$$	$$2$$
$$15$$	$$1$$	$$3$$
$$25$$	$$3$$	$$6$$
$$45$$	$$4$$	$$10$$
$$50$$	$$5$$	$$15$$

Sum of life span $$=2+1+3+4+5\\ =15(odd\quad number)$$

The Median is at $$\left( \cfrac { n+1 }{ 2 } \right) =\left( \cfrac { 15+1 }{ 2 } \right) =8\quad position$$

Now, add up the average life span taken as

position $$=2,2+3=5,3+3=6,6+4=10$$

$$8$$th position comes after $$6$$th position but before $$10$$th position.

So, Median is $$45$$

Question 8
1 / -0

The percentage of marks obtained by the studentsin a class of $$50$$ are given below. Find the median for the following data.
Marks$$(\%)$$ $$40-50$$ $$50-60$$ $$60-70$$ $$70-80$$ $$80-90$$ $$90-100$$
Number of students $$6$$ $$12$$ $$14$$ $$8$$ $$6$$ $$4$$

A
$$65$$

B
$$63.5$$

C
$$64.5$$

D
$$65.5$$

Solution

Marks$$(\%)$$ Number of students(f) Cumulative frequency(cf)
$$40-50$$ $$6$$ $$6$$
$$50-60$$ $$12$$ $$18$$
$$60-70$$ $$14$$ $$32$$
$$70-80$$ $$8$$ $$40$$
$$80-90$$ $$6$$ $$46$$
$$90-100$$ $$4$$ $$50$$
Median is $$25,26$$
Median$$=l+\displaystyle\frac{\frac{n}{2}-cf}{f}\times h$$
where the lower limit,(l) of median class$$= 60$$
the number of observations,$$(n) = 50$$
the cumulative frequency of class preceding the median class,$$(cf) = 18$$
the frequency of median class,$$(f) = 14$$ class size $$(h) = 10$$
Substituting the above values in the formula, we get
Median $$=60+\displaystyle\frac{\frac{50}{2}-18}{14}\times 10$$
$$=60+0.5\times 10$$
$$=65$$
Therefore, the median of the above data is $$65$$

Question 9

1 / -0

The percentage of marks obtained by the students in a class of $$50$$ are given below. Find the mode for the following data.

Marks$$(\%)$$	$$40-50$$	$$50-60$$	$$60-70$$	$$70-80$$	$$80-90$$	$$90-100$$
Number of horses	$$6$$	$$12$$	$$14$$	$$8$$	$$6$$	$$4$$

$$62.5$$

$$63.5$$

$$64.5$$

$$65.5$$

Solution

Marks	Number of Horses
$$40-50$$	$$6$$
$$50-60$$	$$12$$
$$60-70$$	$$14$$
$$70-80$$	$$8$$
$$80-90$$	$$6$$
$$90-100$$	$$4$$

$$Maximum frequency=14$$, Modal class$$=60-70$$

$$l=60$$, $${ f }_{ 1 }=14$$, $${ f }_{ 0 }=12$$, $${ f }_{ 2 }=8$$

$$Mode=l+\left( \cfrac { { f }_{ 1 }-{ f }_{ 0 } }{ { 2f }_{ 1 }-{ f }_{ 0- }{ f }_{ 2 } } \right) \times h\\ \quad =60+\left( \cfrac { 14-12 }{ 28-12-8 } \right) \times 10\\ =60+\left( \cfrac { 20 }{ 8 } \right) =60+2.5\quad \quad \\ =62.5Ans\quad $$

Question 10
1 / -0

The following table of grouped data represents the weight (in kg) of $$100$$ gas cylinders. Calculate the mode weight of a cylinder.
Weight$$(\%)$$ $$3-5$$ $$5-7$$ $$7-9$$ $$9-11$$ $$11-13$$ $$13-15$$
Number of gas cylinders $$16$$ $$13$$ $$15$$ $$25$$ $$14$$ $$17$$

A
$$12.96$$

B
$$15.96$$

C
$$9.96$$

D
$$10.96$$

Solution

The lower limit ($$l$$) of the modal class $$= 9$$,
the class size ($$ h$$) $$= 2,$$
the frequency $$(f_1)$$ of modal class$$ = 25,$$
the frequency $$(f_0 )$$ of the class preceding the modal class$$ = 15$$,
the frequency $$(f_2)$$ of the class succeeding
the modal class $$=14.$$

Now, using the formula:
$$\begin{aligned}{l}{\text{Mode}} &= l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\\ &= 9 + \left( {\frac{{25 - 15}}{{(2 \times 25) - 15 - 14}}} \right) \times 2\\ &= 9 + 0.48 \times 2\\& = 9 + 0.96\\ &= 9.96\end{aligned}$$

Therefore, the mode weight of a cylinder is $$9.96\text{ kg}.$$

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Re-Attempt Weekly Quiz Competition

Farm size	$$12$$	$$13$$	$$14$$	$$15$$	$$17$$	$$18$$	$$19$$	$$20$$
Number of animals	$$3$$	$$4$$	$$4$$	$$2$$	$$1$$	$$2$$	$$4$$	$$1$$

score	$$10$$	$$15$$	$$20$$	$$25$$	$$30$$
frequency	$$2$$	$$3$$	$$5$$	$$6$$	$$4$$

Colour Pencils	$$0-5$$	$$5-10$$	$$10-15$$	$$15-20$$	$$20-25$$	$$25-30$$
Number of students	$$14$$	$$16$$	$$17$$	$$13$$	$$8$$	$$12$$

Weight$$(\%)$$	$$3-5$$	$$5-7$$	$$7-9$$	$$9-11$$	$$11-13$$	$$13-15$$
Number of gas cylinders	$$16$$	$$13$$	$$15$$	$$25$$	$$14$$	$$17$$

Statistics Test - 48

Result

Statistics Test - 48

Score

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Rank

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SHARING IS CARING

Question 1

$$13$$

$$14$$

$$19$$

All the above

Solution

Question 2

$$22$$

$$22.5$$

$$20$$

$$25$$

Solution

Question 3

$$11$$

$$15$$

$$19$$

$$16$$

Solution

Question 4

$$15.25$$

$$12.25$$

$$14.25$$

$$13.25$$

Solution

Question 5

$$65$$

$$32$$

$$34$$

$$31$$

Solution

Question 6

$$69.2$$km

$$70.2$$km

$$72.2$$km

$$73.2$$km

Solution

Question 7

$$10$$

$$15$$

$$25$$

$$45$$

Solution

Question 8

$$65$$

$$63.5$$

$$64.5$$

$$65.5$$

Solution

Question 9

$$62.5$$

$$63.5$$

$$64.5$$

$$65.5$$

Solution

Question 10

$$12.96$$

$$15.96$$

$$9.96$$

$$10.96$$

Solution

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