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BDA 24103- ENGINEERING STATISTICS
TEST 1, Semester 1, Session 2015/2016
Name
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Matrix No.
:.....................................................................................................................
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Lecturer
:...........................................................................................................
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Section
:.................................................................................................
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Instructions: Answer all questions in this exam sheet. (Time: 1
houronly)
Q1 !" Calculate the following probabilities.
(i) P(X !) when X " #($% &.')(ii) P(X ') when X " #(&%
&.*)(iii) P(X + $) when X " #(,% &.$)(i-) P(! X ) when X "
#(/% &./)
( & mar0s )
(a) P(X=3) =
0.2¿¿
P( X =3)= 5!
3! (5−3 ) !¿
(b) P ( X ≤2 )= P ( X =0 )+ P
( X =1 )+ P ( X =2 ) =
10!
0! (10−0 ) ! (0.6
)
0
(1
−0.6
)
10−0
+
10!
1! (10−1 )! (0.6
)
1
(1
−0.6
)
10−1
+
10 !
2 ! (10−2 ) ! (0.6
)
2
(1
−0.6
)
10−2
= 0.0123
GOOD LUCK
#NI$ERSITI T#N %#SSEIN &NN 'A(A)SIA
1acult2 of Mechanical 3 Manufacturing 4ngineering
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(c)
P ( X ≥5 )= P ( x=5 )+ P
( X =6 )+ P ( X =7 )+ P
( X =8)+ P ( X =9 )=¿ 9!
5 ! (9−5 ) !(0.5 )5 (1−0.5)9−5+ 9
!
6 ! (9−6 )!(0
¿0.5000
(d) P (3≤ X ≤ 4 )= P ( X =3 )+ P
( X =4)
¿ 8!
3 ! (8−3 ) !(0.8 )3 (1−0.8 )8−3+ 8
!
4 ! (8−4 )!(0.8 )4 (1−0.8)8−4
¿0.0551
Q2 5he concentration of a reactant is a ran6om -ariable
with probabilit26ensit2
function
f ( x)={1.2 ( x+ x
2
)− x ,∧0
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(b)
mean,μ= E ( X )=∫0
1
x f ( x ) dx=∫0
1
1.2 x3+0.2 x2 dx=
1.2 x4
4+0.2 x
3
3 |=[1,2(1)4
4+0.2(1)3
3 ]−[0 ]=0.3+0.066(c) xiswithin±0.1of the mean
→ (0.3367−0.1 )
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Q3 5he number of hits on a certain website follows a
Poisson6istribution with a
mean rate of / per minute.
(a) 7hat is the probabilit2 that $ messages are recei-e6 in
a
gi-en minute8
(b) 7hat is the probabilit2 that , messages are recei-e6 in
.$minutes8
(c) 7hat is the probabilit2 that fewer than ! messages
arerecei-e6 in a
perio6 of !& secon6s8
( 1& $ar!s )
(a) 'et X be the nu$ber of $essaes received in one $inute.Since
the $ean rate is $essae per $inute X * Poisson()
P ( X =5 )=e−8 85
5 !=0.0916
(b) 'et X be the nu$ber of $essaes received in 1.+ $inute.
Since the $ean rate is $essae per $inute X * Poisson(1,)
P ( X =9 )=e−12 129
9! =0.0874
(c) 'et X be the nu$ber of $essaes received in one half
$inute.
Since the $ean rate is $essae per $inute X * Poisson(-) P
( X
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- End of Question –
GOOD LUCK
5
