DALHOUSIE UNIVERSITY FACULTY OF SCIENCE
Department of Mathematics and Statistics
STATISTICS 2060 Midterm II Examination SOLUTIONS
Date and Time: 9:35-10:55 AM, Monday 15 June, 2009
Name: Student ID #:
The entire exa
...
DALHOUSIE UNIVERSITY FACULTY OF SCIENCE
Department of Mathematics and Statistics
STATISTICS 2060 Midterm II Examination SOLUTIONS
Date and Time: 9:35-10:55 AM, Monday 15 June, 2009
Name: Student ID #:
The entire exam is worth 50 points. It is closed book, but you may refer to one sheet (8.5 x 11in,
both sides) of notes. The exam consists of 6 problems, the number of points allocated to each part of
a problem is shown in the left hand column. You may use a calculator and straight edge.
1. If X is a normal random variable with mean 25 and standard deviation 15, compute the following
probabilities by standardizing.
(a) Compute P(15 ≤ X < 75).(2)
• P(15 ≤ X < 75) = P(
15−25
15 ≤ Z < 75−25
15 ) = P(−0.667 ≤ Z < 3.333)
• P(15 ≤ X < 75) = Φ(3.33) − Φ(−0.67) = 0.9996 − 0.2514 = 0.7482
(b) Compute P(−15 ≤ X < −5).(2)
• P(−15 ≤ X < −5) = P(
−15−25
15 ≤ Z < −5−25
15 ) = P(−2.67 ≤ Z < −2.00)
• P(−15 ≤ X < −5) = Φ(−2.00) − Φ(−2.67) = 0.0228 − 0.0038 = 0.019
2. For Z, the standard normal random variable with mean 0 and standard deviation 1, compute
the following:
(a) Compute the value of c, where P(0 ≤ Z ≤ c) = 0.352.(2)
• P(0 ≤ X ≤ c) = Φ(c) − Φ(0) = 0.352
• Φ(c) = 0.352 + Φ(0) = 0.352 + 0.500 = 0.852
• c = 1.045
(b) Compute the value of c, for P(c ≤ |Z|) = 0.369.(2)
• P(|Z| ≥ c) = 1 − P(|Z| ≤ c) = 1 − P(−c ≤ Z ≤ c) = 0.369
• P(|Z| ≥ c) = 1 − (Φ(c) − (1 − Φ(c))) = 1 − (2Φ(c) − 1) = 0.369
• 2(1 − Φ(c)) = 0.369
• Φ(c) = 1 −
0.369
2 = 0.8155
• c = 0.90
3. During summer months, the weekly demand for Chapman’s ice cream (in 100’s of litres) at the
Mumford Road Sobey’s store is a continuous random variable, X, whose probability density
function is given as:
f(x) = (
1
3
(
x
3
2 − 1) 1.00 ≤ x ≤ 2.46
0 Otherwise
(3) (a) Determine the functional form of the cumulative distribution function, F(x), for rv X.
• F(x) = R x
−∞
1
3
(
y
3
2 − 1)dy =
1
3
h
y
4
8
|
x
1.00 − y|
x
1.00i
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