Statistics Test

Result

Statistics Test - 31

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 following table shows the weight of fifteen students. Find the mode.
Weight (kg) 60 26 53 50 47
Number of students 4 2 2 3 4

A
$$60 kg, 47$$ kg

B
$$60$$ kg

C
$$47$$ kg

D
None of these

Solution

Maximum number of students have weight $$60$$ kg and $$ 47$$ kg. Thus, mode is both $$60$$ kg and $$ 47$$ kg.
Question 2
1 / -0

Find the median weight of the data.
Weight (kg.) $$35$$ $$36$$ $$38$$ $$40$$ $$42$$ $$44$$ $$45$$
Number of students $$6$$ $$5$$ $$8$$ $$9$$ $$2$$ $$7$$ $$4$$

A
$$38$$

B
$$39$$

C
$$40$$

D
$$41$$

Solution

Total number of students $$= 6 + 5 + 8 + 9 + 2 + 7 + 4 = 41$$
Since, the number of students is odd, the median will be $${(\cfrac{n+1}{2})}^{th}$$ number which is the $$21^{st}$$ number
Here, $$21^{st}$$ observation( 4 th row if we create cumulative frequency table) is 40 Kg if arranged in ascending order.
Hence, median $$= 40$$.
Question 3
1 / -0

Find the mode of:
$$x$$
$$15$$
$$16$$ $$17$$
$$18$$
$$19$$
$$20$$ $$21$$
$$22$$
$$23$$
$$f$$
$$6$$
$$7$$
$$9$$
$$13$$
$$10$$
$$12$$ $$8$$
$$0$$
$$4$$

A
$$17$$

B
$$18$$

C
$$20$$

D
$$21$$

Solution

We know that, mode is the observation with the highest frequency.

In the given data, the highest frequency value is $$13$$. This value corresponds to the observation $$18$$.

Hence, the mode of the given data is $$18$$.
Question 4
1 / -0

Find the mode of:
$$7, 7, 8, 10, 10, 11, 10, 13, 14$$

A
$$7$$

B
$$8$$

C
$$10$$

D
$$13$$

Solution

Mode is defined as the maximum number of times an observation appears in the whole set.
Here, by observation, we get that $$10$$ is appearing the maximum (i.e. $$3$$) times and hence this is the mode.
Question 5
1 / -0

Find the mean of the following frequency distribution:
Age (in years)
$$18$$
$$19$$
$$20$$
$$21$$
$$22$$
No. of boys
$$14$$
$$10$$
$$20$$
$$24$$
$$12$$

A
$$19.125$$ years

B
$$20.125$$ years

C
$$21.125$$ years

D
$$21.175$$ years

Solution

Consider the following table to calculate Mean:
$$x_i$$ $$f_i$$ $$x_if_i$$
$$18$$ $$14$$ $$252$$
$$19$$ $$10$$ $$190$$
$$20$$ $$20$$ $$400$$
$$21$$ $$24$$ $$504$$
$$22$$ $$12$$ $$264$$
here,
$$\Sigma f_i=80$$

$$\Sigma f_ix_i=1610$$

$$Mean=\dfrac{\Sigma f_ix_i}{\Sigma f_i}$$

$$\therefore Mean=\dfrac{1610}{80}$$

$$\therefore Mean=20.125$$
Hence, option $$B$$ is correct.
Question 6
1 / -0

Following table gives frequency distribution of electricity consumption of a household in certain area in a month.
No. of units of electricity 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100
No. of households 4 16 47 65 8
Find modal no. of units of electricity consumed by households in a month.

A
$$64$$

B
$$62.8$$

C
$$63.8$$

D
None of these

Solution

Frequency distribution table:
$$ci$$ $$f_i$$
$$0-20$$ $$4$$
$$20-40$$ $$16$$
$$40-60$$ $$47$$
$$60-80$$ $$65$$
$$80-100$$ $$8$$
$$Mode=l+\bigg ( \dfrac{f_1-f_0}{2f_1-f_0-f_2} \bigg )\times h$$
Here,
maximum frequency $$f_1=65$$
$$\Rightarrow$$ modal class $$=60-80$$
lower limit of modal class $$l=60$$
class size $$h=20$$
preceding frequency of modal class $$f_0=47$$
succeeding frequency of modal class $$f_2=8$$
$$\therefore Mode=14.05+\bigg ( \dfrac{65-47}{130-47-8} \bigg )\times 20$$
$$\therefore Mode=14.05+\bigg ( \dfrac{18}{75} \bigg )\times 20$$
$$\therefore Mode=64.8$$
Hence, option $$D$$ is correct.
Question 7
1 / -0

Following are the marks obtained by students in mathematics, draw histogram and hence find the mode
Marks 0 - 19 20 - 39 40 - 59 60 - 79 80 - 99
No.of students 4 8 32 16 8

A
Mode $$\approx$$ $$51$$

B
Mode $$\approx$$ $$41$$

C
Mode $$\approx$$ $$51.5$$

D
Mode $$\approx$$ $$61.5$$
Question 8
1 / -0

Following table shows frequency distribution of no. of rooms occupied in a hotel per day.
No. of rooms occupied $$0 - 10$$ $$10 - 20$$ $$20 - 30$$ $$30 - 40$$ $$40 - 50$$ $$50 - 60$$
No. of days $$5$$ $$27$$ $$17$$ $$11$$ $$9$$ $$1$$
Find median number of rooms occupied per day in a hotel.

A
$$29.76$$

B
$$27.76$$

C
$$23.76$$

D
$$21.76$$

Solution

There are total $$70$$ days. Hence, the median will be the mean of $$35$$ th and $$36$$ th day. Both of which lie in the interval $$20-30$$
The median of grouped data is given by:

Median = $$ L + ( \frac{\frac{n}{2} - cf_{b}}{f_{m}} )\times w$$

where:
L is the lower class boundary of the group containing the median
n is the total number of data
$$cf_{b}$$ is the cumulative frequency of the groups before the median group
$$f_{m}$$ is the frequency of the median group
w is the group width

Hence,
Median =  $$ 20 + ( \frac{\frac{70}{2} - 32}{17} )\times 10$$
Median =  $$ 20 + ( \frac{3}{17} )\times 10$$
Median =  $$ 20 + 1.76$$
Median =  $$ 21.76$$
Question 9
1 / -0

Below is given frequency distribution of driving speed (in kms/hour) of a vehicle of $$400$$ college students.
Speed (in kms/hr) 20 -30 30 - 40 40 - 50 50 - 60 60 - 70
No. of students 6 80 156 98 60
Find modal driving speed of college students.

A
$$41.67$$ kms/hour

B
$$45.67$$ kms/hour

C
$$49.67$$ kms/hour

D
$$53.67$$ kms/hour

Solution

The modal group is $$40 - 50$$.
Estimated Mode $$=$$ $$L + (\frac{f_{m} -f_{m-1}}{(f_{m} -f_{m-1}) + (f_{m} -f_{m+1})}) \times w$$
where:
$$L$$ is the lower class boundary of the modal group
$$f_{m-1}$$ is the frequency of the group before the modal group
$$f_{m}$$ is the frequency of the modal group
$$f_{m+1}$$ is the frequency of the group after the modal group
$$w$$ is the group width
Estimated Mode $$=$$ $$L + \left (\dfrac{f_{m} - f_{m-1}}{(f_{m} -f_{m-1}) + (f_{m} -f_{m+1})}\right) \times w$$
Mode $$=$$ $$40 + \left (\dfrac{156 - 80}{(156 - 80) + (156- 98)}\right)\times 10$$
Mode $$=$$ $$40 +\left (\dfrac{76}{134}\right)\times 10$$
Mode $$= 45.67$$
Question 10
1 / -0

Following table gives frequency distribution of amount of bonus paid to the workers in a certain factory.
Bonus paid (in Rs.) Below 500 Below 600 Below 700 Below 800 Below 900 Below 1000 Below 1100
No. of workers 4 12 24 41 51 58 60
Find median amount of bonus paid to the workers.

A
$$801.27$$ Rs.

B
$$812.27$$ Rs.

C
$$846.27$$ Rs.

D
$$735.29$$ Rs.

Solution

Consider the following table to calculate median:
$$ci$$ $$fi$$ $$cf$$
<500 4 4
500-600 12-4=8 12
600-700 24-12=12 24
700-800 41-24=17 41
800-900 51-41=10 51
900-1000 58-51=7 58
1000-1100 60-58=2 60
Median $$=l +\dfrac {(\dfrac{N}{2}-cf)}{ f_m}\times c$$
Here,
Class Interval $$c=100$$
$$N=\Sigma f_i=60$$
$$\dfrac {N}{2}=30 $$
$$\Rightarrow$$ Median class $$=700-800$$
$$\Rightarrow$$Frequency of median class$$ f_m=17$$
Lower boundary of median class $$l=700$$
previous cumulative frequency of median class $$cf=24$$

$$\therefore$$ Median $$=700+\dfrac {(\dfrac{60}{2}-24)}{ 17}\times 100$$

$$\therefore$$ Median $$=700+\dfrac {(6)}{ 17}\times 100$$

$$\therefore$$ Median $$=735.29$$
Hence. option $$D$$ is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Weight (kg.)	$$35$$	$$36$$	$$38$$	$$40$$	$$42$$	$$44$$	$$45$$
Number of students	$$6$$	$$5$$	$$8$$	$$9$$	$$2$$	$$7$$	$$4$$

$$x$$	$$15$$	$$16$$	$$17$$	$$18$$	$$19$$	$$20$$	$$21$$	$$22$$	$$23$$
$$f$$	$$6$$	$$7$$	$$9$$	$$13$$	$$10$$	$$12$$	$$8$$	$$0$$	$$4$$

Age (in years)	$$18$$	$$19$$	$$20$$	$$21$$	$$22$$
No. of boys	$$14$$	$$10$$	$$20$$	$$24$$	$$12$$

$$x_i$$	$$f_i$$	$$x_if_i$$
$$18$$	$$14$$	$$252$$
$$19$$	$$10$$	$$190$$
$$20$$	$$20$$	$$400$$
$$21$$	$$24$$	$$504$$
$$22$$	$$12$$	$$264$$

No. of units of electricity	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100
No. of households	4	16	47	65	8

$$ci$$	$$f_i$$
$$0-20$$	$$4$$
$$20-40$$	$$16$$
$$40-60$$	$$47$$
$$60-80$$	$$65$$
$$80-100$$	$$8$$

Marks	0 - 19	20 - 39	40 - 59	60 - 79	80 - 99
No.of students	4	8	32	16	8

No. of rooms occupied	$$0 - 10$$	$$10 - 20$$	$$20 - 30$$	$$30 - 40$$	$$40 - 50$$	$$50 - 60$$
No. of days	$$5$$	$$27$$	$$17$$	$$11$$	$$9$$	$$1$$

Speed (in kms/hr)	20 -30	30 - 40	40 - 50	50 - 60	60 - 70
No. of students	6	80	156	98	60

Bonus paid (in Rs.)	Below 500	Below 600	Below 700	Below 800	Below 900	Below 1000	Below 1100
No. of workers	4	12	24	41	51	58	60

$$ci$$	$$fi$$	$$cf$$
<500	4	4
500-600	12-4=8	12
600-700	24-12=12	24
700-800	41-24=17	41
800-900	51-41=10	51
900-1000	58-51=7	58
1000-1100	60-58=2	60

Statistics Test - 31

Result

Statistics Test - 31

Score

-

Rank

-

SHARING IS CARING

Question 1

$$60 kg, 47$$ kg

$$60$$ kg

$$47$$ kg

None of these

Solution

Question 2

$$38$$

$$39$$

$$40$$

$$41$$

Solution

Question 3

$$17$$

$$18$$

$$20$$

$$21$$

Solution

Question 4

$$7$$

$$8$$

$$10$$

$$13$$

Solution

Question 5

$$19.125$$ years

$$20.125$$ years

$$21.125$$ years

$$21.175$$ years

Solution

Question 6

$$64$$

$$62.8$$

$$63.8$$

None of these

Solution

Question 7

Mode $$\approx$$ $$51$$

Mode $$\approx$$ $$41$$

Mode $$\approx$$ $$51.5$$

Mode $$\approx$$ $$61.5$$

Question 8

$$29.76$$

$$27.76$$

$$23.76$$

$$21.76$$

Solution

Question 9

$$41.67$$ kms/hour

$$45.67$$ kms/hour

$$49.67$$ kms/hour

$$53.67$$ kms/hour

Solution

Question 10

$$801.27$$ Rs.

$$812.27$$ Rs.

$$846.27$$ Rs.

$$735.29$$ Rs.

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.