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(MATH144)2007_f_midterm_math144.pdf
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Hong Kong University of Hong Kong Math144 Applied Statistics 2007/2008 Semester 1
Test solution
1. (a) The graph is omitted.
(b) The frequency distribution table is
Class Interval Class Boundaries Class marks Frequency
1 -5 6 -8 9 -11 12 -14 15 -17 18 -20 21 -35 0.5 -5.5 5.5 -8.5 8.5 -11.5 11.5 -14.5 14.5 -17.5 17.5 -20.5 20.5 -25.5 3 7 10 13 16 19 23 150 287 337 403 393 305 125

2 f
n = f =2,000 xf = 26,026 x = 394,058
xf
Mean = . = 13.013
f
nx2 f .( xf)2
Standard deviation = =5.264
n(n .1) 66
Mode = 11.5+ 3 =14.105
66+ 10
1000 .774
Median = 11.5+ 3 =13.182
403
mean -mode 13.013 .14.105
Coe.cient of skewness = = = .0.207
std. dev. 5.264
(c) It is negatively skewed. It skews to the left moderately.
2. (a) i. Let X be the random variable of the number of defective item in the sample. Then X hyp(50,10,0.3).
P(accept) = P(X 1)
= P(X = 0)+ P(X = 1) C15 C35 C15 C35
010 19
=+
C50 C50
10 10
=0.149
1

ii. Let X be the random variable of the number of defective item in the sample. Then X hyp(10,0.16).
P(reject) = P(X> 1)
=1.P(X 1)
=1.P(X = 0) .P(X = 1)
C42 C42
C8 C8
010 19
=1. +
C50 C50
10 10
=0.4919
(b) Let X be the random variable of the number of computer require to repair. Then X Bin(300,0.008). UsingPoisson Approximation (with = np = 2.4),
P(X< 3) = P(X 2)
.2.4 2.40 2.41 2.42
= e ++
0! 1! 2! =0.5697
(c) De.ne S = X1 + X2 .
i. E(X1 )= 0(0.5)+ 1(0.3)+ 2(0.2) =0.7 and Var(X1 )=0.61
ii. The probability distribution of S is
x1 + x2 0 1 2 3 4
P(S = s) 0.25 0.3 0.29 0.12 0.04

3. (a) i. Let C be the civil servant and L be the LegCo.
P(2C 2L) = = = C6 2 C10 2 C16 4 15 45 1820 0.3709
ii.
P(5L)+P(4L1C)+P(3L2C) = C6 0 C10 5 C16 5 + C6 1 C10 4 C16 5 + C6 2 C10 3 C16 5

252+ 1260+ 1800
=
4368 =0.758
(b) X N(5.1,2.12 )
i. P(X> 6) = P(Z> (6 .5.1)/2.1) = P(Z> 0.43) = 0.3336 a .5.1
ii. P(X>a) =0.15 implies = 1.034 =7.2714. He should
2.1
promises 7.27 weeks.
2 4. (a) Let S be the saving account and I be the investment account.
P(I)=0.65 P(S)=0.78 P(S|I)=0.85,
i. P(S I)= P(S|I)P(I)=0.85 . 0.65 = 0.5525.
ii.
P(S I)= P(S)+P(I).P(S I) =0.78+0.65 .0.5525 =0.8775
iii. P(S I).P(S I)=0.8775 .0.5525 =0.325.
(b) Let F be .awed, R be repaired.
P(F)=0.3 P(R|F)=0.75
i.
P(not discard) = 1.P(Rc F) =1.P(Rc |F)P(F) =1.(1 .0.75)(0.3) =1.0.075 =0.925.
0.3 0.75
ii. The required probability = =0.2432
0.925
3