Measures of Central Tendency Test

Result

Measures of Central Tendency Test - 51

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

If Mean deviation about median is $$7.60$$ coefficient of mean deviation about median is $$12.45$$. Find the median____.

A
$$63.75$$

B
$$61$$

C
$$62.21$$

D
$$59.75$$

Solution

Median is the middle most value of a given series that represents the whole class of the series. When mean deviation about the median and its coefficient is given, then
Median = mean deviation about the median/ coefficient of mean deviation about the median
=> Median = 7.60 / 0.1245
= 61
Question 2
1 / -0

Calculate Median salary from the following frequency distribution:

X (Rs. in 000) $$10-20$$ $$20-30$$ $$30-40$$ $$40-50$$ $$50-60$$ $$60-70$$
Y (Frequency) $$2$$ $$3$$ $$6$$ $$5$$ $$2$$ $$2$$

A
$$38.5$$

B
$$36.98$$

C
$$38.33$$

D
$$39.00$$
Solution
Median is the middle most value of a given series that represents the whole class of the series. For a group data,
Median = L + [{(n/2) – B}/G] × w where
L is the lower class boundary of the group containing the median
n is the total number of values
B is the cumulative frequency of the groups before the median group
G is the frequency of the median group
w is the group width
Since the median is the middle value, which in this case is the 10^th one, which is in the 30-40 group. Therefore , 30-40 is the median group so
L = 30
n = 20
B = 2 + 3 = 5
G = 6
w = 10
Median= 30 + [{(20/2) – 5}/6] × 10
= 30 + 8.33
= 38.33
Question 3
1 / -0

The median of $$14, 12, 10, 9, 11$$ is

A
$$11$$

B
$$10$$

C
$$9.5$$

D
$$10.5$$

Solution

Median = Arranging no. in increasing order
$$9,10,11,12,14$$
Median = $$\dfrac{n+1}{2}=\frac{5+1}{2}$$
$$=3rd$$ term $$=11$$
Question 4
1 / -0

Calculate arithmetic mean salary from the following frequency distribution
X(Rs.in 000) 10-20 20-30 30-40 40-50 50- 60 60- 70
Y(Frequency) 2 3 6 5 2 2

A
352

B
36.00

C
35.5

D
39.00

Solution

X 10-20 20-30 30-40 40-50 50-60 60-70
Y 2 3 6 5 5 5
Class interval mid values (x) f d=x-35 fd
10-20 15 2 -20 -40
20-30 25 3 -10 -30
30-40 35 6 0 0
40-50 45 5 10 50
50-60 55 2 20 40
60-70 65 2 30 60
Total 22 80
$$\overline{x} =A+\dfrac{\sum fd}{N}=35+\dfrac{(80)}{32}=38.6$$
Given Cups $$=400$$
$$\therefore 12\%$$ were broken
$$12\%$$ of $$400=\dfrac{12}{100}\times 400=48$$
$$\therefore48$$ cups were broken
In good condition $$=400-48=352$$
Question 5
1 / -0

If x is the average(arithmetic mean) of $$5$$ consecutive odd integers, what is the median of integers?

A
$$0$$

B
$$1$$

C
$$x-2$$

D
x
Question 6
1 / -0

When there are 2 observations in the middle, median is calculated by ______.

A
taking both the values as median

B
taking the mean of the two observations

C
(N+1) / 2

D
both B and C

Solution

Median is the middle most value of a series. So when the series has odd number of elements then median can be calculated easily but when the series has even number of elements then the series has two middle values, so median is calculated either by taking out the average of both the value or the median is the (N+1)/2 th element of the series.
Question 7
1 / -0

In a unimodal and symmetric distribution, the relationship between averages is like this

A
Mean > median > mode

B
Mean < median < mode

C
Mean = median = mode

D
Mean > median < mode
Question 8
1 / -0

For the following grouped frequency distribution find the mode:
Class: 3-6 6-9 9-12 12-15 15-18 18-21 21-24
Frequency: 2 5 10 23 21 12 3

A
13.9

B
14.7

C
15.1

D
14.6
Question 9
1 / -0

List of 5 pulse rates is: 70, 64, 80, 74, 92.what is the median for this list?

A
74

B
76

C
77

D
80

Solution

List of pulse rates: $$64,70,74,80,92$$
Median$$=74$$ (middle value)
Question 10
1 / -0

Find the median of the given ogive:

A
150

B
200

C
148

D
175

Solution

From the given graph, median corresponds to $$300$$ count.
Therefore, median=value of $$x$$ at $$y=300$$
$$=175$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

X (Rs. in 000)	$$10-20$$	$$20-30$$	$$30-40$$	$$40-50$$	$$50-60$$	$$60-70$$
Y (Frequency)	$$2$$	$$3$$	$$6$$	$$5$$	$$2$$	$$2$$

X(Rs.in 000)	10-20	20-30	30-40	40-50	50- 60	60- 70
Y(Frequency)	2	3	6	5	2	2

Class interval	mid values (x)	f	d=x-35	fd
10-20	15	2	-20	-40
20-30	25	3	-10	-30
30-40	35	6	0	0
40-50	45	5	10	50
50-60	55	2	20	40
60-70	65	2	30	60
Total		22		80

Class:	3-6	6-9	9-12	12-15	15-18	18-21	21-24
Frequency:	2	5	10	23	21	12	3

Measures of Central Tendency Test - 51

Result

Measures of Central Tendency Test - 51

Score

-

Rank

-

SHARING IS CARING

Question 1

$$63.75$$

$$61$$

$$62.21$$

$$59.75$$

Solution

Question 2

$$38.5$$

$$36.98$$

$$38.33$$

$$39.00$$

Solution

Question 3

$$11$$

$$10$$

$$9.5$$

$$10.5$$

Solution

Question 4

352

36.00

35.5

39.00

Solution

Question 5

$$0$$

$$1$$

$$x-2$$

x

Question 6

taking both the values as median

taking the mean of the two observations

(N+1) / 2

both B and C

Solution

Question 7

Mean > median > mode

Mean < median < mode

Mean = median = mode

Mean > median < mode

Question 8

13.9

14.7

15.1

14.6

Question 9

74

76

77

80

Solution

Question 10

150

200

148

175

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.