STA301 Assignment No. 1 Fall 2011 solution
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Assignment.1 (Lessons 1-10)
Question 1:Marks: 5+5=10
Age Group (Men) | f |
20-29 | 5 |
30-39 | 4 |
40-49 | 5 |
50-59 | 8 |
60-69 | 12 |
70-79 | 16 |
Total | 50 |
(ii) Write down the following figure in the form of frequency table and also find the relative frequency.
Question 2:Marks: 5+5=10
Height | Number (f) |
57.5-60.0 | 6 |
60.0-62.5 | 26 |
62.5-65.0 | 190 |
65.0-67.5 | 281 |
67.5-70.0 | 412 |
70.0-72.5 | 127 |
72.5-75.0 | 38 |
Total | 1080 |
Statistics Test Scores | |
Stem | Leaf |
6 | 1 1 4 6 7 8 |
7 | 2 3 5 7 9 |
8 | 1 3 5 6 6 7 7 8 9 |
9 | 0 0 3 4 6 8 9 9 |
10 | 0 0 |
a) What is the best test score?
Question 3:Marks: 3+3+4=10(i) Find the median for the following discrete frequency distribution.
Number of pupils per class | Number of Classes |
20 | 1 |
21 | 0 |
22 | 1 |
23 | 3 |
24 | 6 |
25 | 9 |
26 | 8 |
27 | 10 |
28 | 7 |
Total | 45 |
(ii) Calculate the weighted mean from the following data
Item | Expenditure | Weights |
Food | 290 | 7.5 |
Rent | 54 | 2.0 |
Clothing | 98 | 1.5 |
Fuel and light | 75 | 1.0 |
Other items | 75 | 0.5 |
(iii) A man gets a rise of 10% in salary at the end of his first year of service and further of 20 % and 25 % at the end of the second and third year respectively. The rise in each case is calculated on his salary at the beginning of the year. To what annual percentage increase is this equivalent?
Solution:
Class Boundaries f
19.5-29.5 5
29.5-39.5 4
39.5-49.5 5
49.5-59.5 8
59.5-69.5 12
59.5-79.5 16
total 50
(b) How many men fall in the age of 50 or above?
Men above 50
= 8+12+16=36
(c) men fall between the age group of 60 and 69=
=12
(ii) Write down the following figure in the form of frequency table and also find the relative frequency.
Class Boundaries f Relative frequncy
19.5-29.5 5 5/50=0.1
29.5-39.5 4 4/50=0.08
39.5-49.5 5 5/50=0.1
49.5-59.5 8 8/50=
59.5-69.5 12 12/5=
59.5-79.5 16 16/50=
total 50
Question 2:Marks: 5+5=10
(i) Calculate the mode from the following continuous distribution.
Height Number (f)
57.5-60.0 6
60.0-62.5 26
62.5-65.0 190
65.0-67.5 281----f1
l--67.5-70.0 412---model class
70.0-72.5 127-----f2
72.5-75.0 38
Total 1080
Ans=
Mode = l + fm-f1/(fm-f1)+(fm-f2) * h
Fm = frequency of model class = highest frequency =412
H= difference between class boundries = 2.5
So…
Mode= 67.5+412-281/(412-281)+(412-127)*2.5
= 67.5+131/(131)+(285)*2.5
= 68.287
(ii). Use the stem and leaf plot to answer these questions.
Statistics Test Scores
Stem Leaf
6 1 1 4 6 7 8
7 2 3 5 7 9
8 1 3 5 6 6 7 7 8 9
9 0 0 3 4 6 8 9 9
10 0 0
a) What is the best test score?
Ans best score 100
b) How many students take the test?
Ans 30students
c) How many students scored 90?
Ans 02students made 90
d) What is the lowest score?
Ans lowest score= 61
e) Find the difference between the high and low scores.
Ans high score = 100
Lowest = 61
Difference = 100- 61 =59
Question 3:Marks: 3+3+4=10(i) Find the median for the following discrete frequency distribution.Number of pupils per class Number of Classes
F cf
20 1 1
21 0 1
22 1 2
23 3 5
24 6 11
25 9 20
26 8 28
27 10 38
28 7 45
Total 45
Median=n+1/2 for more detail page no61 of HO
45+1/2
median =23rd value is lying in 26the distribution so median = 26
(ii) Calculate the weighted mean from the following dataItem Expenditure Weights
Food 290 7.5
Rent 54 2.0
Clothing 98 1.5
Fuel and light 75 1.0 solve from page no58 of HO
Other items 75 0.5
(iii) A man gets a rise of 10% in salary at the end of his first year of service and further of 20 % and 25 % at the end of the second and third year respectively. The rise in each case is calculated on his salary at the beginning of the year. To what annual percentage increase is this equivalent?
Ans
Annual percentage increase in his salary = 10%+20%+25%/3
= 55/3
= 18.33%