Statistics Test

Result

Statistics Test - 44

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

Find the Arithmetic mean of the above data

$$75.17$$

$$67.17$$

$$76.1$$

$$57.177$$

Solution

By direct method:

Marks Obtained ($$x_i$$)	No. of students ($$f_i$$)	$$f_i x_i$$
$$50$$	$$3$$	$$150$$
$$60$$	$$7$$	$$420$$
$$70$$	$$5$$	$$350$$
$$80$$	$$2$$	$$160$$
$$90$$	$$10$$	$$900$$
$$100$$	$$2$$	$$200$$
	$$\sum f_i = 29$$	$$\sum f_i x_i = 2180$$

$$\overline x=\cfrac{\sum f_i x_i}{\sum f_i} $$

$$= \cfrac{2180}{29}$$

$$ = 75.17$$

Question 2
1 / -0

Identify the model class for the following data$$:$$
$$x$$ $$1-2$$
$$2-3$$
$$3-4$$
$$4-5$$
$$5-6$$
$$f$$ $$0$$
$$4$$
$$3$$
$$2$$
$$1$$

A
$$2-3$$

B
$$1-2$$

C
$$3-4$$

D
$$5-6$$

Solution

Modal class is one, which has the highest frequency.

The highest frequency is $$4$$ which is associated with the class $$2-3$$
Therefore, the modal class is $$2-3$$.
Question 3
1 / -0

A college teacher has the following absentee record of $$50$$ students of a class for the whole year. Find the median.
Number of days
$$0-4$$
$$4-8$$
$$8-12$$
$$12-16$$
$$16-20$$
Number of students
$$10$$
$$18$$
$$9$$
$$1$$
$$12$$

A
$$11.66$$

B
$$12.66$$

C
$$13.66$$

D
$$10.66$$

Question 4

1 / -0

Find the median for the following data shows that distance covered by $$200$$ people to perform their IT project.

Distance(km)	$$5-15$$	$$15-25$$	$$25-35$$	$$35-45$$
Number of people	$$60$$	$$40$$	$$80$$	$$20$$

$$12$$ km

$$13$$ km

$$14$$ km

$$15$$ km

Solution

Distance(km)	Number of people (f)	Cumulative frequency (cf)
$$5-15$$	$$60$$	$$60$$
$$15-25$$	$$40$$	$$60+40=100$$
$$25-35$$	$$80$$	$$100+80=180$$
$$35-45$$	$$20$$	$$180+20=200$$
	$$n=\Sigma f=200$$ $$\displaystyle\frac{n}{2}=\frac{200}{2}=100$$

Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
$$n=\Sigma f=$$ Total number of frequency $$=200$$
$$\displaystyle\frac{n}{2}=\frac{200}{2}=100$$
So, the median class is $$25-35$$
L$$=$$ Lower limit of the median class $$=25$$
cf$$=$$ Cumulative frequency of the class preceding the median class $$=180$$
f$$=$$ Frequency of the median class $$=80$$
h$$=$$ class width$$=10$$
Therefore, Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
$$=25+\left(\displaystyle\frac{\displaystyle\frac{200}{2}-180}{10}\right)\times 10$$
$$=14+\left(\displaystyle\frac{100-180}{80}\right)\times 10$$
$$=25-10$$
$$=15$$
Hence, the median is $$15$$km.

Question 5

1 / -0

Calculate the median of the farm size for the following data$$:$$

Farm size	$$2-5$$	$$5-8$$	$$8-11$$	$$11-14$$	$$14-17$$
Rooms	$$4$$	$$8$$	$$12$$	$$16$$	$$20$$

$$9.125$$

$$8.125$$

$$7.125$$

$$6.125$$

Solution

Farm size	Frequencies(f)	Cumulative frequency(cf)
$$2-5	$$4$$	$$4$$
$$5-8$$	$$8$$	$$4+8=12$$
$$8-11$$	$$12$$	$$12+12=24$$
$$11-14$$	$$16$$	$$24+16=40$$
$$14-17$$	$$20$$	$$40+20=60$$
	$$n=\Sigma f=60$$ $$\displaystyle\frac{n}{2}=\frac{60}{2}=30$$

Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
$$n=\Sigma f=$$ Total number of frequency $$=60$$
$$\displaystyle\frac{n}{2}=\frac{60}{2}=30$$
So, the median class is $$11-14$$
L$$=$$ Lower limit of the median class $$=11$$
cf$$=$$ Cumulative frequency of the class preceding the median class $$=40$$
f$$=$$ Frequency of the median class $$=16$$
h$$=$$ class width$$=3$$
Therefore, Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
$$=11+\left(\displaystyle\frac{\displaystyle\frac{60}{2}-40}{16}\right)\times 3$$
$$=11+\left(\displaystyle\frac{30-40}{16}\right)\times 3$$
$$=11-1.875$$
$$=9.125$$
Hence, the median is $$9.125$$

Question 6
1 / -0

Identify the modal class for the following data $$:$$
Farm
$$20-30$$
$$30-40$$
$$40-50$$
$$50-60$$
$$60-70$$
Animals
$$12$$
$$14$$
$$21$$
$$34$$
$$15$$

A
$$60-70$$

B
$$50-60$$

C
$$30-40$$

D
$$20-30$$

Solution

Modal class is the class which has the highest frequency.
The highest frequency in the given data is $$34$$ which is associated with the class $$50-60$$.
Therefore, the modal class is $$50-60$$.
The answer is $$B$$

Question 7

1 / -0

The following frequency distribution showing the marks obtained by $$100$$ students in Maths. Find the median.

Marks	$$10-20$$	$$20-30$$	$$30-40$$	$$40-50$$
Frequency	$$10$$	$$15$$	$$25$$	$$50$$

$$40$$

$$50$$

$$60$$

$$70$$

Solution

Marks	Frequencies$$(f)$$	Cumulative frequency$$(cf)$$
$$10-20$$	$$10$$	$$0$$
$$20-30$$	$$15$$	$$10+15=25$$
$$30-40$$	$$25$$	$$25+25=50$$
$$40-50$$	$$50$$	$$50+50=100$$
	$$n=\Sigma f=100$$ $$\displaystyle\frac{n}{2}=\frac{100}{2}=50$$

Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
$$n=\Sigma f=$$ Total number of frequency $$=100$$
$$\displaystyle\frac{n}{2}=\frac{100}{2}=50$$
So, the median class is $$40-50$$
$$l=$$ Lower limit of the median class $$=40$$

$$cf$$= Cumulative frequency of the class preceding the median class = $$50$$

$$f=$$ Frequency of the median class $$=50$$
$$h=$$ class width$$=10$$
Therefore, Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
$$=40+\left(\displaystyle\frac{\displaystyle\frac{100}{2}-50}{50}\right)\times 10$$
$$=40+\left(\displaystyle\frac{50-50}{50}\right)\times 10$$
$$=40+0$$
$$=40$$
Hence, the median is $$40$$. (option $$A$$)

Question 8

1 / -0

Estimate the median for the following data$$:$$

Daily wages	$$22-24$$	$$24-26$$	$$26-28$$	$$28-30$$	$$30-32$$
Number of workers	$$23$$	$$120$$	$$100$$	$$110$$	$$105$$

$$27.72$$

$$26.67$$

$$36.67$$

$$66.67$$

Solution

Daily wages

Number of workers(f)

Cumulative frequency (cf)

$$22-24$$

$$23$$

$$24-26$$

$$120$$

$$23+120=143$$

$$26-28$$

$$100$$

$$143+100=243$$

$$28-30$$

$$110$$

$$243+110=353$$

$$30-32$$

$$105$$

$$353+105=458$$

Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$

$$n=\Sigma f=$$ Total number of frequency $$=458$$

$$\displaystyle\frac{n}{2}=\frac{458}{2}=229$$

So, the median class is $$26-28$$

L$$=$$ Lower limit of the median class $$=26$$

cf$$=$$ Cumulative frequency of the class preceding the median class

$$=143$$

f$$=$$ Frequency of the median class $$=100$$

h$$=$$ class width$$=2$$

Therefore, Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$

$$=26+\left(\displaystyle\frac{\displaystyle\frac{458}{2}-143}{100}\right)\times 2$$

$$=26+\left(\displaystyle\frac{229-143}{100}\right)\times 2$$

$$=26+1.72$$

$$=27.72$$

Hence, the median is $$27.72$$.

Question 9

1 / -0

Calculate the median of income used by employee for the following data $$:$$

Income	$$4-14$$	$$14-24$$	$$24-34$$	$$34-44$$
Employee	$$12$$	$$10$$	$$8$$	$$4$$

$$18$$

$$19$$

$$20$$

$$21$$

Solution

Income	Employees(f)	Cumulative frequency (cf)
$$4-14$$	$$12$$	$$12$$
$$14-24$$	$$10$$	$$12+10=22$$
$$24-34$$	$$8$$	$$22+8=30$$
$$34-44$$	$$4$$	$$30+4=34$$
	$$n=\Sigma f=34$$ $$\displaystyle\frac{n}{2}=\frac{34}{2}=17$$

Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
$$n=\Sigma f=$$ Total frequency $$=34$$
$$\displaystyle\frac{n}{2}=\frac{12}{2}=17$$
So, the median class is $$14-24$$
L$$=$$ Lower limit of the median class $$=14$$
cf$$=$$ Cumulative frequency of the class preceding the median class $$=12$$
f$$=$$ Frequency of the median class $$=10$$
h$$=$$ class width$$=10$$
Therefore, Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
$$=14+\left(\displaystyle\frac{\displaystyle\frac{24}{2}-12}{10}\right)\times 10$$
$$=14+\left(\displaystyle\frac{17-12}{10}\right)\times 10$$
$$=14+5$$
$$=19$$
Hence, the median is $$19$$.

Question 10
1 / -0

Find the median for the grouped data given below$$:$$
Marks
$$50-60$$
$$60-70$$
$$70-80$$
$$80-90$$
Students
$$3$$
$$4$$
$$8$$
$$5$$

A
$$53.75$$

B
$$63.75$$

C
$$73.75$$

D
$$83.75$$

Solution

Marks
Students(f)
Cumulative frequency(cf)
$$50-60$$
$$3$$
$$3$$
$$60-70$$
$$4$$
$$3+4=7$$
$$70-80$$
$$8$$
$$7+8=15$$
$$80-90$$
$$5$$
$$15+5=20$$

$$n=\Sigma f=20$$
$$\displaystyle\frac{n}{2}=\frac{20}{2}=10$$
Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
$$n=\Sigma f=$$ Total number of frequency $$=20$$
$$\displaystyle\frac{n}{2}=\frac{20}{2}=10$$
So, the median class is $$70-80$$
L$$=$$ Lower limit of the median class $$=70$$
cf$$=$$ Cumulative frequency of the class preceding the median class $$=15$$
f$$=$$ Frequency of the median class $$=8$$
h$$=$$ class width$$=10$$
Therefore, Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
$$=70+\left(\displaystyle\frac{\displaystyle\frac{20}{2}-15}{8}\right)\times 10$$
$$=70+\left(\displaystyle\frac{10-15}{8}\right)\times 10$$
$$=70-6.25$$
$$=63.75$$
Hence, the median is $$63.75$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

$$x$$	$$1-2$$	$$2-3$$	$$3-4$$	$$4-5$$	$$5-6$$
$$f$$	$$0$$	$$4$$	$$3$$	$$2$$	$$1$$

Number of days	$$0-4$$	$$4-8$$	$$8-12$$	$$12-16$$	$$16-20$$
Number of students	$$10$$	$$18$$	$$9$$	$$1$$	$$12$$

Farm	$$20-30$$	$$30-40$$	$$40-50$$	$$50-60$$	$$60-70$$
Animals	$$12$$	$$14$$	$$21$$	$$34$$	$$15$$

Marks	Students(f)	Cumulative frequency(cf)
$$50-60$$	$$3$$	$$3$$
$$60-70$$	$$4$$	$$3+4=7$$
$$70-80$$	$$8$$	$$7+8=15$$
$$80-90$$	$$5$$	$$15+5=20$$
	$$n=\Sigma f=20$$ $$\displaystyle\frac{n}{2}=\frac{20}{2}=10$$

Statistics Test - 44

Result

Statistics Test - 44

Score

-

Rank

-

SHARING IS CARING

Question 1

$$75.17$$

$$67.17$$

$$76.1$$

$$57.177$$

Solution

Question 2

$$2-3$$

$$1-2$$

$$3-4$$

$$5-6$$

Solution

Question 3

$$11.66$$

$$12.66$$

$$13.66$$

$$10.66$$

Question 4

$$12$$ km

$$13$$ km

$$14$$ km

$$15$$ km

Solution

Question 5

$$9.125$$

$$8.125$$

$$7.125$$

$$6.125$$

Solution

Question 6

$$60-70$$

$$50-60$$

$$30-40$$

$$20-30$$

Solution

Question 7

$$40$$

$$50$$

$$60$$

$$70$$

Solution

Question 8

$$27.72$$

$$26.67$$

$$36.67$$

$$66.67$$

Solution

Question 9

$$18$$

$$19$$

$$20$$

$$21$$

Solution

Question 10

$$53.75$$

$$63.75$$

$$73.75$$

$$83.75$$

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.