Section 4.4 Solutions
4.65
Let G represent the stock gaining 30% and L represent the stock losing 25%.
Furthermore, let the subscripts of 1 and 2 denote the days. Then the sample
space is
.
Let X be the value of the stock after two
days. Then
P(X = $1690) = 
P(X = $975) = 
P(X = $562.50) = 
Hence, the
probability distribution of X is
|
Value of X |
$562.50 |
$975 |
$1690 |
|
Probability |
0.25 |
0.50 |
0.25 |
So the probability
that the stock is worth more after two days than the $1000 I paid for it is
0.25.
The mean/expected
value of the stock after two days is
The mean of $1050.63
tells you the long-run average value of the stock after many, many 2-day
periods.
4.67
a. Let X
be the number of heads in one toss of a fair coin. Then P(X = 0) = 0.5 and P(X
= 1) = 0.5. Hence, 
. Also, 
, so 
So the mean and
standard deviation of the number of heads are 0.5 and 0.5, respectively.
b. Let 
be the number of heads
on the first flip, 
be the number of heads
on the second flip, 
be the number of heads
on the third flip, and 
be the number of heads
on the fourth flip. The number of heads on each of the coin tosses has mean 0.5
and standard deviation 0.5. Furthermore, the coin tosses are independent. Let 
. Then, by the rules for means and variances:
and 
So the mean and standard deviation for the total number of heads in four
flips are 2 and 1, respectively.
c. Based on Example 4.17,
so 
.
These answers agree with part b.
4.69
Let Y be the length of the rod. Then 
and 
. Let be the length of the
left-hand bearing, and let be the length of the
right-hand bearing. Then 
and 
. Let L be the
total length of the assembly.
Then 
. So 
, 
, and 
.
Hence, the mean and
standard deviation of the total length of the assembly are 14 cm and 0.0052 mm,
respectively.
4.70
First we must
convert the standard deviation units to centimeters:
a.
Then by the
Empirical Rule, about 95% of rods have lengths within two standards of 10 cm –
that is, within 2(0.0005) = 0.001 cm of 10. Also, about 95% of bearings have
lengths within 0.0002 cm of 2.
b.
From the
previous problem, the total length has mean 14 cm and standard deviation
0.00052 cm. Hence, 95% of total assemblies have lengths within 2(0.00052) =
0.00104 cm of 14.
4.71
a. In this case, X and Y are not
independent, since knowing the sum of the first two cards will change the
probabilities of the sum of all three cards.
b. In this case, X and Y are independent,
since knowing the sum of the first two rolls will not change the probabilities
of sum of the second two rolls (die rolls are independent).
4.74
Let X be the height, in centimeters, of a
randomly selected 20-year-old man, and Y
be the height, in inches, of a randomly selected 20-year-old man. Then 
. Hence,
and 
. So the mean and standard deviation in inches are 69.61 and
2.83, respectively.
b.
So the insurance company’s mean cash intake is $623.22 per
policy.
.