Measures of Central Tendency Test - 7

N= 5 ,  Values of all observations =  72+86+92+63+77= 390  i.e. 90 (6 x 80) number less  all the observations of  Bob’s score.
Bob need to score in 6th Test = 90
So that his Sum of the values of all 6 test =480
 

When the frequencies are not properly distributed it is called as an asymmetrical or skewed distribution. If it is moderately asymmetrical distribution the following empirical relationship holds good given by Karl Pearson and is expressed as:  Mode = 3 Median - 2 Mean.  144 - 80 = 64

The sum of the deviations from the mean is always zero.

Here, the class 20-30 has maximum frequency. So, it's the modal class. Using interpolation formula of mode, we have 
Mode=l+h (f1-f0)/2f1-f0-f2
=20+10 (27-23)/2×27-23-21
=20+4 =24 
Where, f1 is frequency of modal class, for, is frequency of class just after modal class and f0 is frequency of class preceding modal class. 

Steps involved in calculating median using less than Ogive approach -

1. Convert the series into a 'less than ' cumulative frequency distribution.
2. Let N be the total number of students whose data is given.N will also be the cumulative frequency of the last interval. Find the (N/2) th item(student) and mark it on the y-axis.In this case the (N/2)th item (student) is (60/2) = 30 student.
3. Draw a perpendicular from 30 to the right to cut the Ogive curve at a point 
4. From point where the Ogive curve is cut, draw a perpendicular on the x-axis. The point at which it touches the x-axis will be the median value of the series.

Here, mean =10

X1+X2+X3+X4+X5/5=10

X1+X2+X3+X4+X5=50........................1

Now, mean of three of them =9

X1+X2+X3/3=9

X1+X2+X3=27..................2

Substituting 2 in 1, we get 

27+X4+X5=50 

X4+X5=23

Now, mean of remaining two is 

X4+X5/2=23/2=11.5