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please check your solution again because i think it is going to be wrong, the reason in mistake in your calciulation, the correct answer is 163.621
Khurram bhai ap ka answer 163.621 wrong ha... Asim ka solution teek ha
U can also calculate ur Geometric Mean result from calculator on this side
q2 ma H.M k liye f/x find krna hai answer 164.862 ata hai
in question number 2 the calculation of cumulative frequency mean to say C.F is 190 not 180. so please check your answer before to solve it
STA301_100%_Correct_Solution
Fall2012_STA301_1_SOLUTION
See the attached file please
plz 2nd question ke part 1 ka median part show kr dy plz
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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