STA301 assignment no 1_SOLUTION

[Imran Jafery]

 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 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 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

[mc120200085]

aap ne 1st qs me array type data se class kaise bnai hai plz tell me like 148-152 ye wali??need method??

[Muhammad Afzal]

148 lower limit ha aur upper limit hum is traqiay say find kr sakaty hain. 

upper limit = lower limit +(Class Interval - 1)

=148+(4)

=152

Ab  next upper calss k lia 152 main 5 add krain. aur is tra 5 add krtay jain sari classes pata chal jain gi.

[SANA]

Upper Class limit

148+5=153

153+5=158

158+5=163

163+5=168

168+5=173

Lower Class Limit

152+5=157

157+5=162

162+5=167

167+5=172

172+5=177

now

CLASS BOUNDRIES:

148-152

153-157

158-162

163-167

168-172

173-177

 

[+ M.Tariq Malik]

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 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 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

 

 

[sidra ulfat]

Salam ! there is some confusion in Q2 part (a)...... where you find out the median and you put the values of l, h, f, n, and c are wrong. The correct values will be l = 160, h = 4, f = 58, n = 95, c = 96

actually here first we find the value of n that is n = 190 and than n / 2 = 190/2 = 95

Ater that all the values of l , h ,f, c  will come
after the calculations the answer of the median will be 163.93
Is it so ??? 

[sidra ulfat]

median value will be 159.7

[adeel ghaffar]

bhai plzzz tel me Q 2 k B part harmonic mean mia 1/x   find kare ge mean X kis ko consider  kare ge 

[umair]

 

Q No.1 (a)

 

 

Classes                        F          cumulative Frequency                     

 

 

148-152                       6                                  6                                 

 

153 -  157                    11                                17                               

 

158-    162                   14                                31

 

163 -      167                9                                  40

 

168-     172                  3                                  43

 

173     - 177                 2                                  45

 

                                    45

 

Q No.1 (a)

 

 

Classes                        F                       Relative frequency

 

 

148 - 152                     6                                  6/45 =0.1333                          

 

153   - 157                   11                                11/45= 0.244                                      

 

158 -   162                   14                                14/45= 0.311

 

163   -      167              9                                  9/45=    0.2

 

168 -    172                  3                                  3/45=   0.066

 

173     - 177                 2                                  2/45= 0.044

 

                                    45