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(MATH244)midterm01F_L2sol.pdf
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Mid-term Examination Solution
1.
(a)
5*|6
6+|34 n = 20
6*|679 leaf unit = 0.1 (100kWh)
7+|00234 * for 0-4
7*|566799 + for 5-9
8+|224

(b)
Negatively skewed.

(c)
Min = 5.6 , Max = 8.4


For QL , p =0.25 , (n +1)p =5.25,
QL =X ()5 +0.25(X () .X 5 )=6.7 +0.25(6.9 .6.7)=6.75

6 ()
For median, p =0.5 , (n +1)p =10.5, meidan=X ()10 +0.5(X ( ) .X 10 )=7.3 +0.5(7.4 .7.3)=7.35
11 ()
For QU, p =0.75 , (n +1)p =15.75, QU =X ()+0.75(X ( ) .X 15 )=7.7 +0.75(7.9 .7.7)=7.85
15 16 ()
Hence the five number summary is : 5.6, 6.75, 7.35, 7.85, 8.4 .
(d) Boxplot for the data :

5.5 6.0 6.5 7.0 7.5 8.0 8.5
2. Let X be the power demand in a particular day. Then X ~ N(15,4).
.17 .15 .
1 0.1587
(a) Pr(X >17)=1..
.=1.()=1.0.8413 =
. 4 .


(b)
Let Y be the number of HPD days in six months. Then Y ~ b(180,0.1587).


. .( )( ).
29.5 180 0.1587
Pr(Y 30)1.. .
=1.(0.19)=1.0.5753 =0.4247
.( )( 180 0.1587)( 0.8413)..
.
(c) Let Xi , i =1,2,...,180 be the power demand in each of the coming 180 days. Then
. 4 .
X~ N.15, .. . 180 .
180
Total revenue = 0.015Xi =2.7X million dollars.
i=1
. 40 .
Pr(Total revenue >40 million dollars)=Pr(2.7X >40)=Pr.X >. . 2.7 .
. ..
.=1.(.1.242)=(1.242)=0.8929
=1.. .. .

P. 1
Mid-term Examination Solution
2.
(d) If the normal assumption is violated, the calculation in (a) will be incorrect. The calculation in (b) relies on the result from (a) and thus is also incorrect. However, the calculation in (c) is still valid because by central limit theorem, the sample mean of a random sample will follow a normal distribution approximately when the sample size is large.

3.
Pr()I 0.3 Pr II =0.3 Pr III 0.4 Pr(defective | I )=0.04 Pr(defective | II )=0.03 Pr(defective | III )=0.02 Pr(defective)=0.04 0.3 +0.030.3 +0.02 0.4 =0.029


Pr(defective | I ) () Pr I 0.04 0.3 12

= ()()=
(i)
Pr(I | defective)= ===0.4138

Pr(defective) 0.029 29
Pr(defective | II ) () Pr II 0.030.3 9


(ii)
Pr(II | defective)= ===0.3103


Pr(defective) 0.029 29
Pr(defective | III )( ) Pr III 0.02 0.4 8

(iii) Pr(III | defective)= ===0.2759
Pr(defective) 0.029 29
4. Let X be the time (in minutes) for a student to finish the examination. Then X (),.~
()
2

(a) E()X =90 . =90 Var X =15 . =225
2
() 90

0.4 EX 0.4 90 36
= EX = = =()= =
() 225
Var X
Hence X ~ (36,0.4).

(b) Let c be the time that should be set. Then Pr(X c)=0.9 .
2
Since 0.8X ~ (36,0.5) ,
72 2
Pr(0.8X 0.8c)=Pr(72 0.8c)=0.9 .0.8c 85.53 .c 107 min (Or, 0.8c85.53 +0.2(96.58 .85.53)=87.74 .c 110 min)
.0.8 .72 . 0.8 .72 1.282 .0.8c 87.384 .c 110 min)
5. Joint and marginal distributions of X and Y:
Y

X 0 1 2 Marginal
0 0.90 0.03 0.02 0.95
1 0.03 0.01 0.01 0.05
Marginal 0.93 0.04 0.03 1

(a) Pr(X0 Pr Y 0)0.95 0.93 0.8835 090 Pr X 0,Y 0)
=)( ===