In: Statistics and Probability
Chapter 2 Exercise is divided in to 2 sections A and B. Data for this assignment is under Data files in module Ex2-30-e8.xls (for A) and Ex2-34-e8.xls (for B). See data under modules.
A) The following data give the weekly amounts spent on groceries for a sample of households.
| 
 $271  | 
 $363  | 
 $159  | 
 $ 76  | 
 $227  | 
 $337  | 
 $295  | 
 $319  | 
 $250  | 
| 
 279  | 
 205  | 
 279  | 
 266  | 
 199  | 
 177  | 
 162  | 
 232  | 
 303  | 
| 
 192  | 
 181  | 
 321  | 
 309  | 
 246  | 
 278  | 
 50  | 
 41  | 
 335  | 
| 
 116  | 
 100  | 
 151  | 
 240  | 
 474  | 
 297  | 
 170  | 
 188  | 
 320  | 
| 
 429  | 
 294  | 
 570  | 
 342  | 
 279  | 
 235  | 
 434  | 
 123  | 
 325  | 
| 
 312  | 
 2,753  | 
 2,595  | 
 6,057  | 
 7,624  | 
 6,624  | 
 6,362  | 
 6,575  | 
 7,760  | 
 7,085  | 
 7,272  | 
| 
 5,967  | 
 5,256  | 
 6,160  | 
 6,238  | 
 6,709  | 
 7,193  | 
 5,631  | 
 6,490  | 
 6,682  | 
 7,829  | 
 7,091  | 
| 
 6,871  | 
 6,230  | 
 7,253  | 
 5,507  | 
 5,676  | 
 6,974  | 
 6,915  | 
 4,999  | 
 5,689  | 
 6,143  | 
 7,086  | 
Solution
Let x = amount ($) spent on groceries
Back-up Theory
Let xi =mid-point and fi be the frequency of ith classi. Then,
Mean (Average), µ, = Σ(i = 1, n)(xi.ri)/Σ(i = 1, n)(fi) ………………………….......................………. (1)
Variance, σ2 = Σ(i = 1, n){fi.(xi – µ)2}/ Σ(i = 1, n)(fi) or equivalently [{Σ(i = 1, n){fi.(xi)2} – µ2 ]...... . (2)
Standard deviation (SD), σ = sqrt(Variance) ……………………………………........................….. (3)
Now to work out the solution,
Final answers are given below. Back-up Theory and Details of calculations follow at the end.
Part (a)
Frequency Distribution
Maximum value: 570
Minimum value: 41
Range of values: 529
Stipulated number of classes: 9
Class width = 529/9 = 59 ~ 60
| 
 Class # i  | 
 Class Boundaries  | 
 Frequency  | 
| 
 1  | 
 40 to 100  | 
 3  | 
| 
 2  | 
 100 to 160  | 
 4  | 
| 
 3  | 
 160 to 220  | 
 6  | 
| 
 4  | 
 220 to 280  | 
 12  | 
| 
 5  | 
 280 to 340  | 
 10  | 
| 
 6  | 
 340 to 400  | 
 2  | 
| 
 7  | 
 400 to 460  | 
 1  | 
| 
 8  | 
 460 to 520  | 
 1  | 
| 
 9  | 
 520 to 580  | 
 1  | 
| 
 Total  | 
 40  | 
Answer 1
Part (b)
Vide (1), mean = 254.5 Answer 2
[Details follow at the end after Part (c).]
Part (c)
Vide (2), variance = 10,509.7500 Answer 3
Vide (3), standard deviation = 102.52 Answer 4
[Details follow]
| 
 Class # i  | 
 Class Boundaries  | 
 Frequency fi  | 
 Mid-point xi  | 
 fi.xi  | 
 di = xi - mean  | 
 di2  | 
 di2.fi  | 
| 
 1  | 
 40 to 100  | 
 3  | 
 70  | 
 210  | 
 -184.5  | 
 34040.25  | 
 102120.75  | 
| 
 2  | 
 100 to 160  | 
 4  | 
 130  | 
 520  | 
 -124.5  | 
 15500.25  | 
 62001  | 
| 
 3  | 
 160 to 220  | 
 6  | 
 190  | 
 1140  | 
 -64.5  | 
 4160.25  | 
 24961.5  | 
| 
 4  | 
 220 to 280  | 
 12  | 
 250  | 
 3000  | 
 -4.5  | 
 20.25  | 
 243  | 
| 
 5  | 
 280 to 340  | 
 10  | 
 310  | 
 3100  | 
 55.5  | 
 3080.25  | 
 30802.5  | 
| 
 6  | 
 340 to 400  | 
 2  | 
 370  | 
 740  | 
 115.5  | 
 13340.25  | 
 26680.5  | 
| 
 7  | 
 400 to 460  | 
 1  | 
 430  | 
 430  | 
 175.5  | 
 30800.25  | 
 30800.25  | 
| 
 8  | 
 460 to 520  | 
 1  | 
 490  | 
 490  | 
 235.5  | 
 55460.25  | 
 55460.25  | 
| 
 9  | 
 520 to 580  | 
 1  | 
 550  | 
 550  | 
 295.5  | 
 87320.25  | 
 87320.25  | 
| 
 Total  | 
 40  | 
 10180  | 
 243722.25  | 
 420390.00  | 
| 
 Mean = 10180/40  | 
 254.5  | 
| 
 Variance = 420390/40  | 
 10,509.7500  | 
| 
 Standerd deviation = √6093.0563  | 
 102.52  | 
DONE