STA301 Assignment%232
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Transcript of STA301 Assignment%232
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8/6/2019 STA301 Assignment%232
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3
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Assignment No.2 (Course STA301)
Spring 2011 (Total Marks 30)
Question 1: Marks: 2+2+4+2=10a. By using five number summary, suggest the skewness of the data from the
following information:
Xm 55,X0 25,Q2 39
Solution: Q2 X0 39 25 14
XmQ2 55 39 16
Distance between Q2 -X0 is less than Xm -Q2,
so it is positively skewed
b. For a certain data,
m1 0, m2 2.64, m3
5.6
and m4 28.30
Find moment ratios b1 and b2 . Interpret your results also.
Solution:MOMENTS RATIOS
m
2
b
and...b m
41
m 3 2 m
22 2
5.62
b and...b
28.302
1
2.64 2.64 31.36 28.30
b and b
1 18.399...
2
6.969
b1=1.70 and b2=4.06 answer
INTERPRETATION OF b1Our distribution is positively skewed, m3 will be positive and our distribution is
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2
2
negatively skewed, m3 will be negative. b m3
INTERPRETAION OF b2
For a normal distribution, b2=3
1
m 3
c.Explain equally likely events with the help of an example.
Answer: two events A and B are said to be equally likely, when one event is as likely
to occur as the other.
Example: When a fair coin is tossed, the head is as likely to appear as the tail, and theproportion of times each side is expected to
appear is 1/2.
Question 2: Marks: 4+4+2=10
a. For a particular
data of 5 pair ofvalues:
Y2
2 6 ,
Y 1 0 ,
X Y 3 7
The fitted regression line is:
Y= -1.5 + 0.5 x
Find the standard error of estimate (Sy.x); what doest it indicates?
Solution:
syx
Y2 aYbXY
n 226 1.5 10 0.537
5 2
7.5
=2.73 answer
b. Five yellow balls and four green balls, which are indistinguishable apart from
color, are placed in a bag if six balls are taken from the bag; Find the probability
of being three yellow and three green.
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Solution :
n !
n1!.n2!.n3!.........
6!
3!X3!6X5X4X3X2X1
6X6
72036
=20 answer
c. If the equations of the least square regression lines are:
Y= -1 + 0.5 x And X= -1.1 + 0.2 y
Find correlation coefficient r.Solution : r
0.31Answer
byxb
xy
0.50.2 0.1
Question 3: Marks: 5+5=10
a.
i. How many different words (even meaningless) can be formed from the word
Component?
Solution:C comes ones so n1= 1
O comes twice n2=2
M comes ones so n3=1
P comes ones so n4=1N comes twice n5=2
E comes ones so n6=1
T comes ones so n7=1
n!
n1 !.n2 !.n3 !.n4 !.n5 !.n6 !.n7 !
9!
1!2!1!1!2!1!1!
=90720answer
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ii. How many distinct four-digit numbers can be performed from the following
integers 2, 3,4,5,6 where each integer is used only once?
Solution:
n1=5
n2=4
n3=3
n4=2
5X4X3X2=120 answer
b. When a pair of dice is rolled, make the sample space and find the probabilityi) Total of 5
Solution :
SAMPLE SPACE{(1.1), (1.2), (1.3), (1.4), (1.5), (1.6)
(2.1), (2.2), (2.3), (2.4), (2.5), (2.6)
(3.1), (3.2, (3.3, (3.4), (3.5), (3.6)
(4.1), (4.2), (4.3), (4.4), (4.5), (4.6)
(5.1), (5.2), (5.3), (5.4), (5.5), (5.6)
(6.1), (6.2),(6.3), (6.4), (6.5), (6.6)}
PROBABILITY OF TOTAL 5 ISn(A)=(1.4 , 2.3, 3.2,4.1,)n(S)=36
n(A)
n(S)4
36
=0.11 answer
ii) Total of 7?
PROBABILITY OF TOTAL 7=?
Let B be the event that sum of outcome is equal to 7
n(A)={1.6, 2.5, 3.4, 4.3, 5.2, 6.1}
n(A)=6
n(S)=36
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n(A)
n(S)
6
36
0.16answer