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2 b)
median=X childa =l+h/f(n/2C.F)
is ma
l= the lower class boundary of the medain class(n/2)
h=the class intrval size of the median class
f = the class frequncy of tmedian class
n=sigma f=total frequncy
C.F=the cumulative frequenchy preceding the medin class cumulativ freq
c.b f c.f
145150 4 4
150155 6 10
155160 28 38
160165 58 96
165170 64 160
170175 30 190
toal 190
median class or the modal class = (n/2)th=190/2=95 th (to see in c.f)
so the medain r modal class is 160165 58 96
by putng the valus in formula
medain = 160 +5/58(9538)
=160+5/58(57)
=164.9138
One more Solution
STA301_Assignment#01_Solution_(www.vustudents.ning.com)
One more idea
STA301_Solved_Assignment#01
Question 1: Marks: 5+2=7
a) Following are the heights (in centimeters) of 45 female high school students. Prepare a frequency distribution of the data where the class width is 5, the number of classes is 6 and the smallest observation is 148. Also obtain the relative frequency and relative cumulative frequency.
170, 151, 154, 160, 158, 154, 171, 156, 160, 157, 160, 157, 148, 165, 158, 159, 155, 151, 152, 161, 156, 164, 156, 163, 174, 153, 170, 149, 166, 154, 166, 160, 160, 161, 154, 163, 164, 160, 148, 162, 167, 165, 158, 158, 176.
ANSWER:
Classes 
(f) 
r.f 
r.c.f 
148  152 
6 
6/45 = ans 
0.133 
153  157 
11 
11/45 =ans 
0.377 
158  162 
14 
14/45 = ans 
0.688 
163  167 
9 
9/45 = ans 
0.888 
168  172 
3 
3/45 =ans 
0.955 
173  177 
2 
2/45 = ans 
0.999 

45 


b) Which level of measurement (scale of measurement) is suitable for the following data in each example?
i) The lake SaifulMuluk is 10,578 feet above from the sea level. (Ratio Scale)
ii) If customers go to the shop for buying shoes and try different size of shoes. The sizes of shoes are 7B, 10D and 12E. (Interval Scale)
iii) The numbering on Tshirts of players in the cricket team. (Ordinal Scale)
iv) The weight of the person is 68 kg. (Nominal Scale)
Question 2: Marks: 5+3=8
a) Find the median, harmonic mean and 23^{rd} percentile for the following frequency distribution.
Class Boundaries 
Frequency 
145 – 150 
4 
150 – 155 
6 
155 – 160 
28 
160 – 165 
58 
165 – 170 
64 
170 – 175 
30 
Solution:
Boundaries 
f 
c.f 
145 – 150 
4 
4 
150 – 155 
6 
10 
155 – 160 
28 
38 
160 – 165 
58 
96 
165 – 170 
64 
160 
170 – 175 
30 
190 
median=l + h/f(n/2− c)
l = 164,h = 5,f = 64,n = 190,c = 96
CALCULATE YOURSELF PLEASE
Class Boundaries 
Frequency 
X 
1/X 
F(1/X) 
145 – 150 
4 
(145 + 150)/2 
0.007 
0.03 
150 – 155 
6 
 
0.007 
0.04 
155 – 160 
28 
 
0.006 
0.2 
160 – 165 
58 
 
0.006 
0.4 
165 – 170 
64 
 
0.006 
0.4 
170 – 175 
30 
 
0.006 
0.2 

190 


TOTAL= 
H.M =n/{∑f(1/x)}
CALCULATE YOURSELF PLEASE
boundaries 
f 
c.f 
145 – 150 
4 
4 
150 – 155 
6 
10 
155 – 160 
28 
38 
160 – 165 
58 
96 
165 – 170 
64 
160 
170 – 175 
30 
190 
P23 = l + h/f(23n/100 − c)
l=164,h = 5,f = 64,n = 190,c = 96
CALCULATE YOURSELF PLEASE
b) Explain the concept of Geometric mean? Also find the geometric mean for the following data.
18, 19, 19, 19, 19, 20, 20, 20, 20, 20, 21, 21, 21, 21, 22, 23, 24, 27, 30, 36
Solution:
x 
logx 
18 
1.26 
19 
1.27 
19 
1.27 
19 
1.27 
19 
1.27 
20 
1.30 
20 
1.30 
20 
1.30 
20 
1.30 
20 
1.30 
21 
1.32 
21 
1.32 
21 
1.32 
21 
1.32 
22 
1.34 
23 
1.36 
24 
1.38 
27 
1.43 
30 
1.48 
36 
1.56 

19.46 
g.m = log G = (∑ log X)/n
CALCULATE YOURSELF PLEASE
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