HE1004 Introduction to Statistical Theory & Methods
Semester 2, 2020-21
Tutorial Exercise No. 7 - Solution
Question 1
In a group of adult males, their uric acid readings in excess of the standard
value 6; is a random variable X with the following pdf
8 1
1<x<1
< 2 (2x 3x2 );
f (x) =
:
0
elsewhere
Calculate the mean excess for this group of adult males.
Solution
By de…nition R
1
E(X) =
xf (x)dx
1
=
=
=
R1
1
2
2 (2x
1
1
2
2 3
3x
3x3 )dx
3 4 x=1
4 x x= 1
2
3
Question 2
The time elapsed, in minutes, between the placement of an order of pizza and
its delivery is random with the pdf
8 1
< 15 ; 25 < x < 40
f (x) =
:
0
elsewhere
(a) Determine the mean and standard deviation of the time it takes for the
pizza shop to deliver pizza.
(b) Suppose it takes 12 minutes for the pizza shop to bake pizza. Determine
the mean and standard deviation of the time it takes for the delivery person to
deliver pizza.
Solution
(a)Let X be the time between the placement of an order and its delivery. Then
R 40 x
E(X) = 25 15
dx = 32:5
E(X 2 ) =
R 40 x2
dx
25 15
= 1075
Therefore, V (X) = 1075
(32:5)2 = 18:75 and hence
1
X
= 4:33
(b) The time it takes for the delivery person to deliver the pizza is X
E(X
X 12
12) = E(X)
=
X
12 = 32:5
12:
12 = 20:5
= 4:33
Question 3
Let X be a random variable with density function
jxj
e
f (x) =
2
;
1<x<1
Find P ( 2 < X < 1)
Solution
P ( 2 < X < 1)
=
=
=
=
R1
R0
R0
1
2
e
jxj
2
2
e
jxj
2
2
e
2
(
dx
x)
2
0
dx +
x=0
[ex ]x=
R1e
dx +
2
+
1
2
=
1
2
1
e
2
+
=
1
2 (2
e
2
e
=
0:7484
jxj
R1e
0
[ e
1
2
1
1
dx
2
x
2
dx
x x=1
]x=0
e
1
)
Question 4
Let X be a continuous random variable with pdf f (x) = 6x(1 x); 0 < x < 1:
What is the probability that X is within 2 standard deviations from the mean?
Solution
2
E(X)
=
=
=
E(X 2 )
=
=
V (X)
P
1
2
p
(2)( 0:05) < X <
1
2
p
+ (2)( 0:05)
R1
0
6x2 (1
1
2
R1
0
6x3 (1
=
3
10
=
0:05
=
P(
=
x)dx
6 5 x=1
5 x x=0
3 4
2x
3
10
=
3 4 x=1
2 x x=0
2x3
=
=
x)dx
1 2
2
p
5 2 5
10
R 0:94721
<X<
0:05279
6x(1
3x2
2x3
p
5+2 5
10
x)dx
x=0:94721
x=0:05279
0:9838
Question 5
It takes a professor a random time between 20 and 27 minutes to walk from
his home to school everyday. If he has a class at 9:00 am and he leaves home
at 8:37 am, …nd the probability that he reaches his class on time
Solution
To reach his class on time, he has to arrive at school in not more than 23
minutes. Let X represent the time he takes to walk to school, then
X R(20; 27):
3
20
P (X 23) = 23
27 20 = 7
Question 6
A line of length ` is divided by a point selected at random within the line.
What is the probability that it will divide the line such that the longer
segment is at least twice the length of the shorter segment?
Solution
The line will be divided into two segments with the longer segment at least
twice the length of the shorter segment i¤ the selected point lies within 13 of
the distance from the left end of the line, or more than 32 of the distance from
the left end of the line. Required probability is therefore 23 :
3
Question 7
The volume of a sphere with radius r is given by V = 43 r3 : The radius of a
sphere is a random number uniformly distributed over the interval [2; 4].
(a) What is the expected value of its its volume?
(b) What is the probability that its volume is at most 36 ?
Solution
(a) E(V ) =
Now,
E(r3 ) =
E(r3 )
R 4 r3
2 2
=
h
=
44
8
=
30
So, E(V ) =
(b)
P (V
4
3
4
3
36 )
r4
8
dr
ir=4
r=2
24
8
30 = 40
4
3
=
P
=
P (r3
=
P (r
=
3 2
4 2
=
1
2
r3
36
27)
3)
Question 8
Suppose that lifetimes of light bulbs produced by a certain company are
normal random variables with mean 1000 hours and standard deviation 100
hours. Is this company correct when it claims that 95% of its light bulbs last
at least 900 hours?
Solution
Let X represent the lifetime of light bulbs produced. Then
X N ( X = 1000; 2X = 1002 )
900 1000
P (X 900) = P Z
= P (Z
1) = 0:8413
100
No, the claim is incorrect. Only about 84% of the light bulbs produced have
lifetimes of at least 900 hours.
Question 9
4
At an archaeological site 130 skeletons are found and their heights are
measured and found to be approximately normal with mean 172 centimeters
and variance 81 centimeters. At a nearby site, …ve skeletons are discovered and
it is found that the heights of exactly three of them are above 185 centimeters.
Based on this information is it reasonable to assume that the second group of
skeletons belongs to the same family as the …rst group of skeletons?
Solution
Let X represent the heights of the …rst group of skeletons. Then
X N ( X = 172; 2X = 81):
P (X > 185) = P Z > 185 9 172 = P (Z > 1:44) = 0:0749
If the second group of skeletons belongs to the same family as the …rst group,
the chance of observing 3 or more skeletons above 185 centimeters is
5
5
5
3
2
4
1
5
0
3 (0:0749) (0:9251) + 4 (0:0749) (0:9251) + 5 (0:0749) (0:9251) = 0:0037
Since this probability is small, it is unlikely that the second group belongs to
the …rst group.
5