Math 170S Homework 1 Solution
Chuyin Jiang
1. 1.
1
x̄ = 30
30
P
xi ≈ 7.2
i=1
1
s2 = 30−1
30
P
(xi − x̄)2 ≈ 4.2
i=1
s ≈ 2.0
2.
24 observations lie within one standard deviation from the mean, 29 lie within
two standard deviations from the mean.
2. Mean:
n
1X
ȳ =
yi
n i=1
n
=
1X
(axi + b)
n i=1
n
1X
1
=a
xi + bn
n i=1
n
= ax̄ + b
1
Variance:
s2y =
=
n
1 X
(yi − ȳ)2
n − 1 i=1
n
1 X
(axi + b − ax̄ − b)2
n − 1 i=1
n
1 X 2
=
a (xi − x̄)2
n − 1 i=1
n
1 X
=a
(xi − x̄)2
n − 1 i=1
2
= a2 s2x
3. 1. First we denote y1 , y2 , . . . , y30 as the sorted data.
π̃0.25 : (30 + 1)0.25 = 7.75 = 7 + 0.75. So π̃0.25 = y7 + 0.75(y8 − y7 ) = 5.975.
π̃0.75 : (30 + 1)0.25 = 23.25 = 23 + 0.25. So π̃0.75 = y23 + 0.75(y24 − y23 ) = 8.325.
2.
π̃0.1 : (30 + 1)0.1 = 3.1 = 3 + 0.1. So π̃0.1 = y3 + 0.1(y4 − y3 ) = 5.02.
π̃0.9 : (30 + 1)0.9 = 27.9 = 27 + 0.9. So π̃0.9 = y27 + 0.9(y28 − y27 ) = 10.46.
3. Suspected outlier is 14.1. And there is no outliers.
4. 1. Recall that 70th percentile means P [xi < π0.7 ] = 0.7.
Y7 < 27.3 means either: All Xi ’s are below 27.3; or: seven of them are below 27.3,
and one is greater than 27.3.
Hence P (Y7 < 27.3) = 87 0.77 × 0.3 + 0.78 .
2. Y5 < 27.3 < Y8 means the number of observations below 27.3 can be 5,6, or 7.
7
P
8
P (Y5 < 27.3 < Y8 ) =
(0.7)i (0.3)8−i .
i
i=5
5. The pdf for U 0, 1 is f (x) = 1, and cdf is F (x) = x. Recall from class FWr (w) =
n
n
k
P
P
n
n
k
n−k
P (Wr ≤ w) =
F
(w)
[1
−
F
(w)]
f
(w)
=
w [1 − w]n−k .
k
k
k=r
k=r
n!
Then we have pdf fWr (w) = (r−1)!(n−r)!
wr−1 (1 − w)n−r
R 1 r+1
R1
R1
n!
E[Wr2 ] = 0 w2 fW (w)dw = 0 = (r−1)!(n−r)!
w (1 − w)n−r = Γ(r+2)Γ(n−r+1)
=
Γ(n+3)
0
(r+1)!(n−r)!
r(r+1)
n!
= (n+1)(n+2)
(r−1)!(n−r)!
(n+2)!
2
r(r+1)
r(n−r+1)
r
2. V ar(Wr ) = E(Wr2 ) − E(Wr )2 = (n+1)(n+2)
= (n+1)
2 = (n+1)2 (n+2) .
Please let me know of any typo/errors, thx:)
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