if u find any mistake so plzzzzz tell
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One more Solution
STA301_Assignment#01_Solution_(www.vustudents.ning.com)
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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 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:
a) Find the median, harmonic mean and 23rd 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:
23rd 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/number-conversion/anti-log-logarithm-calcul...
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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