We have been working very hard since 2009 to facilitate in your learning Read More. We can't keep up without your support. Donate Now.

www.vustudents.ning.com

 www.bit.ly/vucodes + Link For Assignments, GDBs & Online Quizzes Solution www.bit.ly/papersvu + Link For Past Papers, Solved MCQs, Short Notes & More

Looking for Something at Site? Search Below

ENJOY IT...:)

+ http://bit.ly/vucodes (Link for Assignments, GDBs & Online Quizzes Solution)

+ http://bit.ly/papersvu (Link for Past Papers, Solved MCQs, Short Notes & More)

Views: 1017

Attachments:

### Replies to This Discussion

Question 1:                                                                                                                   Marks: 5+2=7

a)   Following are the heights (in centimeters) of 45 female high school studentsPrepare 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.

 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 Saif-ul-Muluk 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 T-shirts 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 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

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

 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)}

 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

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

D= noooooo not fair i made it so please use my name by the way i posted it in the group also

Ðarker Than Black ma nay main discussion say iuthi hai ..islihay lihay ap above dekhu tu ap ki discussion ka link nichay a raha hai

bacho wali assignment thi is pa larny wali kia bat thi wo b full solved nhi hy sara kam tu srudent ko karna hy har jga likha hy...............CALCULATE YOURSELF PLEASE

Fall2012_STA301_1_SOLUTION

Attachments:

aap ne is ki freq kaise bnai hai???????

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.

 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 Saif-ul-Muluk 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 T-shirts 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 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

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

 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)}

 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

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

## HELP SUPPORT

This is a member-supported website. Your contribution is greatly appreciated!

## Latest Activity

+!!! Slow poison!!! liked Shanzay's discussion Kyaa Khuda B Ussi Ka Tha....
1 minute ago
2 minutes ago
5 minutes ago
15 minutes ago
+ıllıllı ᎶᏋᏁᎥᏬᏕ ᏰᎥᏒᎴ ıllıllı+ posted discussions
15 minutes ago
18 minutes ago
19 minutes ago
20 minutes ago

1

2

3