Assignment No: 1 (Lessons 1-9)
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.
b) Which level of measurement (scale of measurement) is suitable for the following data in each example?
i) The lake Saif-ul-Muluk is 10,578 feet above from the sea level.
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.
iii) The numbering on T-shirts of players in the cricket team.
iv) The weight of the person is 68 kg.
Question 2: Marks: 5+3=8
a) Find the median, harmonic mean and 23rd 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 |
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
Tags:
aur ban gai hai :) chk on page 3
secnd qstion k a part ma ap ko medin ka formula dekan ha
median=X childa =l+h/f(n/2-C.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
145-150 4 4
150-155 6 10
155-160 28 38
160-165 58 96
165-170 64 160
170-175 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 160-165 58 96
by putng the valus in formula
medain = 160 +5/58(95-38)
=160+5/58(57)
=164.9138
No it's wrong! mein ny upload key hain file page 3 pr chk karo.
2 b)
median=X childa =l+h/f(n/2-C.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
145-150 4 4
150-155 6 10
155-160 28 38
160-165 58 96
165-170 64 160
170-175 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 160-165 58 96
by putng the valus in formula
medain = 160 +5/58(95-38)
=160+5/58(57)
=164.9138
Q2 part1
median =164.9138
H.M.=164.862
P23=160.49
Q2 Partb
G.M.=21.66
match with yr answers and justify
Galat hai. CF k column mein 96+64 karna hai 160 banay ga aur nichy walaq 190.
One more Solution
STA301_Assignment#01_Solution_(www.vustudents.ning.com)
One more idea
STA301_Solved_Assignment#01
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