Measure of position: A measure of position is a method by which the position that a particular data value has within a a given data set can be identified.
OR
Measure of position are used to locate the relative position of
the data value in the data set.
The most common measure of position are quartiles, deciles,
percentiles and standard score()z score .
Suppose you have a data set 10,20,30,40,50,60
suppose mean =50.
consider any particular value say 10 check its relative position with mean.
Q1-First quartile or (lower quartile)
Q2-Median
Q3-Third quartile or (upper quartile)
Quartiles: Divides the data in four equal parts.
Deciles: Divides the data in ten equal parts.
Percentiles: Divides the data in 100 equal parts.
3 Quartiles
Q1:-lower quartile is the value below which 25% obs’s lie.
Q1=(N+1)/4th item/obs’s.
Q2:Median is the value below which 50 obs’s lie
Q2=2(n+1)/4 th item/obs’s.
Q3:Median is the value below which 50 obs’s lie
Q3= 3(n+1)/4 th item/obs’s
For grouped Data
QK = K(N+1)/4 th term.
Q1 = l+(n/4-cf)/f*i
Q2 = l+(2N/4-cf)/f*i
Q3 = l+(3N/4-cf)/f*i
QK = l+(KN/4-cf)/f*I
9 Deciles:
D1 ,D2 , D3 …… D9 which divides the data into 10 equal parts .for
discrete freq. dist. When arranged in ascending order.
Dk = kth deciles
Dk=k(N+1/10)th item/obs.
For grouped data
Dk=l+(kN/10-c.f/f)*i)
K=1,2,3….9
99 Percentiles:
99 partition (position) values Will divide the data into 100 equal
parts. Percentiles are denoted by p1,p2,p3….p99.
Kth percentiles, Pk=k(N+1/100)TH item/obs.
When data is arranged in ascending order.
For grouped Data
Pk=l+(k.N/100-c.f)/f)*i
Ques.1-Find Q1,Q3,D7,P67
15,10,25,18,17,21,27,29,30
Sol.-Count the number.=9
Arrange the data in ascending order
10,15,17, 18, 21,25,27,29,30
N=9
Q1=(N+1)/4 th item
=(9+1)/4
=10/4
=2.5
2nd item+.5(3rd item-2nd item)
15+.5(17-15)
=15+.5(2)
16
Q3 =(3N+1)/4 th item
=3(9+1)/4
=7.5
=7TH item+ .5(8th -7th item)
=27+.5(29-27)
=28
D7= 7(N+1)/10th term
D7=7(9+1/10)
=7
P67=67(n+1)/100th term
P67=67(9+1)/100
=67*10/100
=6.7
Ques.2-Find Q1,Q3,D7,P67
15,10,25,18,17,21,27,29,30
Sol.-Count the number.=9
Arrange the data in ascending order
10,15,17, 18, 21,25,27,29,30
N=9
Q1=(N+1)/4 th item
=(9+1)/4
=10/4
=2.5
2nd item+.5(4th item-3rd item)
15+.5(18-17)
=15+.5
15.5
Q3 =(3N+1)/4 th item
=3(9+1)/4
=7.5
=7TH item+ .5(8th -7th item)
=27+.5(29-27)
=28
D7= 7(N+1)/10th term
D7=7(9+1/10)
=7
P67=67(n+1)/100th term
P67=67(9+1)/100
=67*10/100
=6.7
=25+.7(27-25)
=25+.7(2)
=26.4
Ques.3 Calculate the lower and upper quartile ,fourth deciles and 60 the percentiles from the following distribution.
Marks group |
No. of student |
|
05-10 |
5 |
|
10-15 |
6 |
|
15-20 |
15 |
|
20-25 |
10 |
|
25-30 |
5 |
|
30-35 |
4 |
|
35-40 |
2 |
|
40-45 |
2 |
Sol:- The cumulative frequency table is
| C.I | f | c.f |
| 05-10 | 5 | 5 |
| 10-15 | 6 | 11 |
| 15-20 | 15 | 26 |
| 20-25 | 10 | 36 |
| 25-30 | 5 | 41 |
| 30-35 | 4 | 45 |
| 35-40 | 2 | 47 |
| 40-45 | 2 | 49 |
Here n=49, N/4= 12.25th term lies in 15-20 class
l=15, cf=11, f=15, i=20-15=5
Q1=l+(N/4-cf)/f*i
=15+(12.25-11)/15*5
=15.41
ii)Q3=3/4N=36.75th term lies in 25-30 class
l=25,cf=36, f=5, i=5.
Q3= l+(3N/4-cf)/f*i
=25+(36.75-36)/5*5
=25.75
iii)D4=4N/10=4/10*49
=19.6th term lies in the
class 15-20.
l=15,cf=11 , f=15,i=5
D4= l5+(4/10N-cf)/f*i
=15+((19.6-11)/15)*5
=17.92
iv)P60=60/100N=60/100*49=29.4th term lies in 20-25
l=20, cf=26, f=10, i=5
P60=l+((60/100N-cf)/f)*i
=20+(29.4-26)/10*5
=20+1.7
=21.7
Ques.4 The scores on a midterm examination are presented
below in decreasing order of magnitude
Find the percentile rank of a score of 63.
65,65,65,64,64,63,62,62,62,61,61,61,60,60,60,60,59,59,58,58,57,57,57,57,57,56,56,
56,48,47
Sol. We know that a percentile rank of k means, k % student has got
marks below pk.
Here pk= 63
No of student whose marks are below 63.
Percentile rank=24/30*100=80%
It means 80% student got marks below Pk.
Standard score or (z score)
Standard score or (z score)-: A z-score is also known as
standard score. Z-score are expressed in term of standard deviation from the
mean.
-If a z-score is equal to zero, it is on the mean.
-if a z-score is equal to +1, it is one S.D above the mean.
-if a z-score is equal to+2, it is 2 S.D above the mean.
Example: Suppose you got 75 marks ,average of the class is 50.
you have
+-σ =68..27%
+-2σ=95.45%
+-3σ =97.73%
In order to use a z-score ,you need to know
the mean ,μ and also the population s.d.
Formula of z -score
z=value-mean/s.d
Example: Raj score 85 in
chemistry, while the class average was 80,with s.d of 5.vineet score 70 in his
class, while average i was 60 with s.d 6. Who score better to their class.
Sol- For Raj
Z=x-μ /σ
z=85-80/5
z=5/5=1σ
For Vineet
z=70-60/6
=10/6
z=1.67σ
means vineet
performed better than raj.
Example:-GMAT average score is 600 with S.D of
250.Find the standard score of person who scores 750.
Sol.-Z=x-μ /σ
750-600/250
=150/200
=.60
The score is .605 above the mean
It means person not performed good.
Z score may also be =+ve and-ve,
With a +ve value
indicating the score is above the mean and -ve score indicate it is below the
score
Example:
Lets x=400, μ =500, σ 50
then Z=x-μ /σ
400-500/50=
-100/50
-2σ
Score is 2σ
below mean
The
performance is 95% bad from the mean .we always measure this in term
of σ not in term of no.
A z-score gives you an idea of how far from the mean a data pt.
is. But more technically it is a measure how many s.d,'s below or above the
population mean a raw score is.
Ex: Knowing that someone's wt. is 85 kg might
be good information, but if you want to compare it to the
"avg" person's weight,
A z-score can tell you where that person's weight is compared to the
avg population's mean weight.
Example:- Lets say
you have a test score of 190
the test has μ =150
σ=25
Z=x-μ /σ
your z score=190-150/25=40/25
=1.6σ
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