STAT-201
FINAL EXAM (KEY)
Question 1
a) Define Population.
A population consists of all subjects that are being studied
SEMESTER-II, 2012-2013
(10 Marks)
1
b) The data shown below are the number of grams per serving of 20 selected brands of cakes.
i ) Construct a grouped frequency distribution using 6 classes.
ii) Draw an ogive of the distribution using relative frequencies.
22
18
12
7
22
16
14
21
23
26
11
25
20
15
27
19
24
16
17
20
SOLUTIONS:
i) Maximum value = 27,
Minimum Value = 7
Class width =
½
1+1+1½ +1
Frequency Distribution
Class Width Class Boundaries Tally
7 – 10
6.5 – 10.5
|
Frequency
1
11 – 14
10. 5 – 14.5
|||
3
15 – 18
14.5 – 18.5
||||
5
19 – 22
18.5 – 22.5
|||| |
6
23 – 26
22.5 – 26.5
||||
4
27 – 30
26.5 – 30.5
|
1
Total
20
ii) Graph
For table: 1, For axes: ½ + ½ , Graph: 2
cumulatie relatie frequency
Ogive
1.00
0.80
0.60
0.40
0.20
0.00
2.5
6.5
10.5
14.5
18.5
Class boundaries
22.5
26.5
30.5
Less than 6.5
Less than 10.5
Less than 14.5
Less than 18.5
Less than 22.5
Less than 26.5
Less than 30.5
c.f
0
1
4
9
15
19
20
c.r.f
0.00
0.05
0.20
0.45
0.75
0.95
1.00
STAT-201
FINAL EXAM (KEY)
SEMESTER-II, 2012-2013
Question 2 ( 7 + 3 )
(10 Marks)
a) The frequency distribution represents the data obtained from a sample of 80 washing machine
service technicians. The values are the days between service calls for various washing machines.
Find: i) Mean
ii) Modal class
iii) Variance
Period(days)
25.5 – 28.5 28.5 – 31.5 31.5 – 34.5 34.5 – 37.5 37.5 - 40.5 40.5 - 43.5
Number of calls
5
9
32
20
12
2
b) The average number of days construction workers miss per year is 11 with standard deviation 5.29 and
the average number of days factory workers miss per year is 8 with standard deviation 3.24. Which
class of workers is more variable in terms of days missed?
SOLUTIONS:
a)
Calculation Table:
Period(days)
25.5 – 28.5
28.5 – 31.5
31.5 – 34.5
34.5 – 37.5
37.5 – 40.5
40.5 – 43.5
Frequency Midpoints
fxm
fxm2
f
xm
5
27
135 3645
9
30
270 8100
32
33
1056 34848
20
36
720 25920
12
39
468 18252
2
42
84 3528
n = 80
2733 94293
1+1+1
1
ii) Modal class = 31.5 – 34.5 days
b) C. Var. formula:
1
½
C. Var. for construction workers =
1
C. Var. for factory workers =
1
Since the coefficient of variation is larger for construction workers, construction workers are more
variable than factory workers.
½
STAT-201
FINAL EXAM (KEY)
SEMESTER-II, 2012-2013
Question 3
a) Find the median of : 60, 70, 30, 20, 10, 50
(10 Marks)
b) Find the boundaries of : 62.5cm
c) Classify as discrete or continuous variable: Lifetimes (in hours) of 25 iPod batteries.
d) Classify as qualitative or quantitative variable: Marital status of nurses in a hospital.
e) Find the complement of the event: Tossing a coin twice and not getting all heads.
f) Of Ph.D. students, 60% have paid assistantships. If 3 are selected at random, find the probability
that none has an assistantship?
g) Using the sample space for tossing two dice, construct the probability distribution for the sums 2
through 12.
SOLUTIONS:
a) Arranging the data:
10, 20, 30, 50, 60, 70
Median = (30 +50) / 2 = 40
1
b) Boundaries of 62.5cm = 62.45 – 62.55 cm.
1
c) Continuous variable.
1
d) Qualitative variable.
1
e) Complement of the event not getting all heads = getting all heads = {HH}
1
f) p(getting assistantships) = 0.6 , p(not getting assistantships) = 1 – 0.6 = 0.4
(1 + 1)
p(none has an assistant ship ) = 0.4 x 0.4 x 0.4 = 0.064
g) S= {(1,1), (1,2), (1,3), ......................................(6,6) }
n(S) = 36
1
Probability distribution:
2
x
2
3
4
5
6
7
8
9
10
11
12
P(x) 1/36 1/18 1/12 1/9 5/36 1/6 5/36 1/9 1/12 1/18 1/36
STAT-201
FINAL EXAM (KEY)
SEMESTER-II, 2012-2013
Question 4
(2 + 6 + 2)
(10 Marks)
a) Find the probability of selecting 3 science books and 4 math books from 8 science books and 9 math
books. The books are selected at random.
b) A sample of 500 respondents were selected in a large city to study consumer behaviour. Among the
questions asked was “Do you enjoy shopping for clothing?” Of 240 males, 136 answered yes. Of 260
females, 224 answered yes. What is the probability that a respondent chosen at random
i) is a female and enjoys shopping for clothing?
ii) is a female or enjoys shopping for clothing?
iii) is a male, given that the respondent answered “No” ?
c)
Find the correlation coefficient between X and Y from the following values:
n = 6,
ΣX = 362,
ΣY = 287,
ΣXY = 18355,
ΣX2 = 22648,
ΣY2 = 15311
SOLUTIONS:
a) Let A be the event of selecting “3 science books and 4 math books”
Then n(A) = 8C3 x 9C4 = 56 x 126 = 7056,
P(A) = 7056 / 19448 = 0.3628
b) Table for calculation:
Yes No
Males
136 104
Females 224 36
Total
360 140
n(S) = 17C7 = 19448
(½ + ½)
1
( Table: 1)
Total
240
260
500
i) P( a female and enjoys shopping) = 224 / 500 = 0.448
1
ii) P(a female or enjoys shopping) =P(a female)+P(enjoys shopping) - P(a female and enjoys shopping)
= 260 /500 + 360 /500 - 224/500 = 396 / 500 = 0.792
(1 + 1)
iii) P(a male, given that the respondent answered NO) = P(a male / answered NO)
(1+ ½ + ½ )
= P(a male and answered NO) / P(answered NO)
= (104/500) / (140/500)
= 104 / 140 = 0.7428
c) Correlation coefficient :
( 1 + 1)
STAT-201
FINAL EXAM (KEY)
SEMESTER-II, 2012-2013
Question 5 (1 + 1+ 3 + 5)
a) Define regression.
(10 Marks)
b) Construct the LCL of an x –control chart if :
= 31.1,
= 17.8, A5 = 0.577, and D5=2.115
c) Seventy randomly selected light bulbs were tested to determine their lifetimes. The following
frequency distribution was obtained. Find the failure and survival probabilities at various points of
the time interval 100 – 150 hours.
Lifetime ‘t’ (hours)
100 – 110 110 – 120 120 – 130 130 – 140
041 - 051
# bulbs failed (Freq.)
8
15
20
15
12
d) A researcher claims that the average cost of men’s athletic shoes is less than $80. He selects a
random sample of 36 pairs of shoes from a catalogue and finds that the mean cost is $75. If the
standard deviation of the population is 19.2, is there enough evidence to support the researcher’s
claim at α = 0.10? [critical value -1.28]
SOLUTIONS:
a) Regression is a statistical method used to describe the nature of the relationship between variable
and to make prediction about the values of the variables.
1
b) LCL =
c)
- A5
= 31.1 - (0.577 x 17.8) = 20.8294
Table of required probabilities
1
½ +1 +1+ ½
Time to failure # of bulbs failed up to t Failure probability Survival probability
‘t’ (hours)
(cumulative frequency)
F(t)
R(t) = 1 – F(t)
100
0
0
1
110
8
0.11
0.89
120
23
0.33
0.67
130
43
0.61
0.39
140
58
0.83
0.17
150
70
1
0
d) Step-1:
Hypothesis:
H0 : μ = 80
H1 : μ ˂ 80
1
Step-2:
Critical value = - 1.28
Step-3:
Computation:
Step-4:
Decision: Since the test value, -1.56 is less than the critical value (falls in the critical
Region) the null hypothesis is rejected.
1
Step-5:
Conclusion: There is a enough evidence to support the claim that the average cost of
the men’s athletic shoes is less than $80.
1
2