Download Solutions to Quiz # 3 (STA 4032)

Solutions to Quiz # 3 (STA 4032)
1. Suppose that the current measuements in a strip of wire are assumed to follow a
normal distribution with mean of 12 milliamperes and a variance of 9 (milliamperes)2 .
(a) What is the probability that a measurement will exceed 15 milliamperes?
(b) What is the probability that a current measurement is between 9 and 15 milliamperes?
(c) Determine the value for which the probability that a current measurement is below this value is 0.95.
Solution. (a) Let X denote the current in milliamperes. The, X has a normal
distribution with µ = 12 and σ = 3. So, the desired probability is given by
15 − 12
X − 12
>
) = P (Z > 1) = 1 − P (Z ≤ 1)
3
3
= 1 − 0.841345 = 0.158655.
P (X > 15) = P (
(b)
9 − 12
X − 12
15 − 12
<
<
) = P (−1 < Z < 1)
3
3
3
= P (Z < 1) − P (Z ≤ −1) = 0.841345 − 0.158655 = 0.68269.
P (9 < X < 15) = P (
(c) We want to find the value of x such that P (X < x) = 0.95. Note that
P (X < x) = P (
X − 12
x − 12
x − 12
<
) = P (Z <
) = 0.95.
3
3
3
Using Table III, we have P (Z < 1.64) = 0.95. So, letting
x = 16.92 milliamperes.
x−12
3
= 1.64, we find
2. The distance between major cracks in a highway follows an exponential distribution with a mean of 5 miles.
(a) What is the probability that there are no major cracks in a 15-mile stretch of the
highway?
(b)What is the probability that the first major crack occurs between 15 and 20 miles
of the start of inspection?
(c) Given that there are no cracks in the first 5 miles inspected, what is the probability that there are no major cracks in the next 20 miles?
Solution. Let X denote the distance between major cracks. Then, X is an exponential random variable with λ = 1/E(X) = 1/5 = 0.2 cracks/mile.
(a)
Z ∞
∞
P (X > 15) =
0.2e−0.2x dx = −e−0.2x = e−3 = 0.04985.
15
15
(b)
Z
20
P (15 < X < 20) =
15
20
0.2e−0.2x dx = e−0.2x 15
= e−3 − e−4 = 0.04985 − 0.01831 = 0.03154.
(c) By the memoryless property, we have
Z ∞
∞
P (X > 20+5 X > 5) = P (X > 20) =
0.2e−0.2x dx = −e−0.2x 20 = e−4 = 0.01831.
20