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STA301_Assignment#01_Solution_(www.vustudents.ning.com)
One more idea
STA301_Solved_Assignment#01
plz measurement waly ka ans bta dain plz plz plz plz
Jazak Allah
Ash Dear ummmah
Class Number 
class limit 
Frequency 
Relative Frequency (r.f) 
Commulative frequency (c.f) 
Relative comulative frequency r.f * c.f=r.cf 
1 
148  152 
6 
6/45 = 0.133 
6 
0.789 
2 
153  157 
11 
11/45 = 0.244 
17 
4.148 
3 
158  162 
14 
14/45 = 0.311 
31 
9.641 
4 
163  167 
9 
9/45 = 0.2 
40 
8 
5 
168  172 
3 
3/45 = 0.066 
43 
2.838 
6 
173  177 
2 
2/45 = 0.044 
45 
1.98 


Total = 45 



y relative cumulative frequency kasy ai?
yar ap ne relative cumulative frequency galat nikali hai. is k lie formula ye hai: c.f/total frequency.
yaha total frequency 45 hai. for example: first relative frequency:6/45=0.133 second:17/45=0.377 third:31/45=0.688 and so on.
Question 1:
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.
Ans:
Smallest value(X0) = 148
Largest value (xm) = 176
Range = Xm – X0 = 176 – 148 = 28
Class interval = h = Range/ No of classes
= 28 / 6 = 4.6 ~ 5
Class Number 
class limit 
Frequency 
Relative Frequency 
Relative cumulative frequency 
1 
148 152 
6 
6/45 = 0.133 
0.133 
2 
153 – 157 
11 
11/45 = 0.244 
0.133+0.244 = 0.377 
3 
158 – 162 
14 
14/45 = 0.311 
0.377+0.311=0.688 
4 
163 – 167 
9 
9/45 = 0.2 
0.688+0.2=0.888 
5 
168 – 172 
3 
3/45 = 0.066 
0.888+0.066=0.954 
6 
173  177 
2 
2/45 = 0.044 
0.954+0.044=0.998 


Total = 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.
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 Tshirts of players in the cricket team.
iv) The weight of the person is 68 kg.
Question 2:
a) Find the median, harmonic mean and 23^{rd} percentile for the following frequency distribution.
Class Boundaries 
Frequency 
CF 
X 
1/x 
F(1/x) 
145 – 150 
4 
4 
147.5 
0.0067 
0.0268 
150 – 155 
6 
10 
152.5 
0.0065 
0.039 
155 – 160 
28 
38 
157.5 
0.0063 
0.1764 
160 – 165 
58 
96 
162.5 
0.0061 
0.3538 
165 – 170 
64 
160 
167.5 
0.0059 
0.3776 
170 – 175 
30 
190 
172.5 
0.0057 
0.171 






Median:
Plz calculate it
Harmonic mean:
23^{rd} Percentile:
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
Ans:
Geometric mean G of a set of n positive values X1, X2…Xn is defined as the positive nth root of their product.
X 
logX 
18 
1.2552 
19 
1.2787 
19 
1.2787 
19 
12.787 
19 
1.2787 
20 
1.3010 
20 
1.3010 
20 
1.3010 
20 
1.3010 
20 
1.3010 
21 
1.3222 
21 
1.3222 
21 
1.3222 
21 
1.3222 
22 
1.3424 
23 
1.3617 
24 
1.3802 
27 
1.4313 
30 
1.4771 
36 
1.5563 

∑logX = 
G = antilog 1.33564 = 21.65
U can also calculat ur result from this calculator on side
U can also calculate Ur ANTILOG from online calculator on the side
http://ncalculators.com/numberconversion/antiloglogarithmcalcul...
please pay attention the formula of harmonic mean is not correct because it is used in ungrouped data as we know that in the question we have provided group data so please resolve your question with correct mean to say with grouped formula
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