Download Exam 3: Solution

The University of Texas at San Antonio
Department of Management Science and Statistics
STA 3513, Probability and Statistics
Instructor: Victor De Oliveira
November 27, 2006
Name:
EXAM 3: SOLUTION
1. The breakdown voltage of a randomly chosen diode of a certain type is known to be normally distributed with mean 40 volts (v) and standard deviation 1.5 v.
(a) What is the probability that the voltage of a single diode is between 39 and 42 ?
(2 points)
(b) What value is such that 15% of all diodes have voltages exceeding that value ?
(2 points)
(c) If four diodes are independently selected, what is the probability that at least one has a voltage
exceeding 42 ?
(3 points)
Answer.
42−40
(a) P (39 < X < 42) = P ( 39−40
1.5 < Z < 1.5 ) = Φ(1.33) − Φ(−0.67) = 0.9082 − 0.2514 = 0.6568 .
(b) value = 85th percentile of the N (40, 1.52 ) distribution, so value = 40 + (1.5)(1.04) = 41.56 .
(c) P(at least one of the diodes has voltage exceeding 42) = 1 – P(all diodes have voltage below 42)
= 1 − (0.9082)4 [by independence] = 0.3197, since P (X ≤ 42) = Φ(1.33) = 0.9082.
2. A pharmaceutical company knows that 5% of its birth-control pills have an ingredient that is below
the minimum strength, thus rendering the pill ineffective. What is the (approximate) probability that
in a sample of 200 pills fewer than 10 will be ineffective ?
(3 points)
Answer. Let X = number of ineffective pills in a sample of 200 ∼ Bin(200, 0.05), so
E(X) = (200)(0.05) = 10 and var(X) = (200)(0.05)(0.95) = 9.5.
Using the normal approximation to the binomial distribution:
P (X < 10) = P (X ≤ 9) ≈ Φ
9 + 0.5 − 10 √
= Φ(−0.1622) ≈ 0.4364
9.5
3. A component has lifetime X (in hours) that is exponentially distributed with parameter β = 25.
(a) Compute the probability that the lifetime of the component surpasses 10 hours.
(3 points)
(b) If the cost of operation per unit time is c = 5, what is the expected cost of operating this component
over its lifetime ?
(2 points)
Answer.
R∞ 1 −x
(a) P (X > 10) = 10 25
e 25 dx = 0.67
(b) Cost = 5X so E(Cost) = 5E(X) = 5 × 25 = 125
2
STA 3513, Probability and Statistics
EXAM 3: SOLUTION
4. In a certain city the daily consumption of water (in millions of liters) follows a gamma distribution
with parameters α = 2 and β = 3.
(a) What is the city expected daily consumption of water ?
(2 points)
(b) If the daily capacity of that city is 9 million liters of water, what is the probability that on a given
day the water supply is inadequate ?
(3 points)
Answer.
X = daily consumption of water ∼ Gamma(2, 3).
(a) E(X) = (2)(3) = 6.
(b) P(water supply is inadequate) = P (X > 9) = 1 − P (X ≤ 9) = 1 − F IG ( 39 ; 2) = 1 − F IG (3; 2)
= 1 − 0.8010 = 0.199.
5. Let X1 , . . . , X49 be a random sample from a distribution with mean µ = 11 and variance σ 2 = 4.
Find the mean and variance of the sample mean X̄.
(2 points)
Answer.
E(X̄) = E(X1 ) = 11,
var(X̄) =
var(X1 )
n
=
4
49 .
6. The breaking strength of a rivet has a mean value of 10000 psi and a standard deviation of 500 psi.
(a) What is the (approximate) probability that the sample mean breaking strength for a random
sample of 40 rivets is between 9900 and 10200 psi ?
(3 points)
(b) What breaking strength value is such that the sample mean of the random sample in part (a)
exceeds that value for (approximately) 5% of the time.
(2 points)
Answer.
approx.
2
By the central limit theorem, X̄ ∼ N (10000, 500
40 ).
9900−10000
10200−10000
√
(a) P (9900 ≤ X̄ ≤ 10200) ≈ P ( 500/ 40 ≤ Z ≤ 500/√40 ) = Φ(2.53) − Φ(−1.26) = 0.8905 .
√
40(a−10000)
a−10000
√
≈ 0.95,
(b) 0.05 = P (X̄ > a) = 1 − P (X̄ ≤ a) ≈ 1 − Φ 500/
⇐⇒
Φ
500
40
so from the standard normal table
√
40(a−10000)
500
= 1.645, and a = 10000 +
1.645(500)
√
40
= 10130.05 .