The algebraic sum of the deviations of a frequency distribution from its mean is always,
>0
<0
0
a non-zero number
Which of the following can not be determined graphically?
Mean
Median
Mode
None
The absccissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its
Mean
Median
Mode
None
20
30
40
50
Mean of 100 items is 49. It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50 respectively. The correct mean is
48
49
50
60
While computing mean of grouped data, we assume that the frequencies are
evenly distributed over all the classes
centred at the classmarks of the classes
centred at the upper limits of the classes
centred at the lower limits of the classes
Mode and mean of a data are 12k and 15A. Median of the data is
12k
14k
15k
16k
If mean = (3 median – mode) . k, then the value of k is
1
2
1/2
3/2
The median of set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 2, then the median of the new set
is increased by 2
is decreased by 2
is two times of the original number
Remains the same as that of the original set.
The median from the table is
Value Frequency 7 2 8 1 9 4 10 5 11 6 12 1 13 3
11
10
12
11.5
Cumulative frequency curve is also called
Histogram
ogive
Bar graph
median
Find the value of x, if the mode of the following data is 25. 15, 20, 25, 18, 14, 15, 25, 15, 18, 16, 20. 25, 20, x, 18
20
25
15
18
If the mode of a data is 18 and the mean is 24. then median is
22
20
18
16
6
7
5
8
20 - 30
30 - 40
10 - 20
None
0.04, 0, 0.04, 0.5
0.04, 0, 0.04, 0.6
0.04, 0, 0.04, 0.2
0.04, 0, 0.04, 0.098
Between 20 and 22.5
Between 27.5 and 30
Between 22 and 27.5 but not equal to 25
Equal to 25
There are lottery tickets labelled numbers from 1 to 500. I want to find the number which is most common in the lottery tickets. What quantity do I need to use?
