Soal Probstat

BAB 2

2-70. Consider the endothermic reactions in Exercise 2-50.

Let A denote the event that a reaction final temperature is

271 K or less. Let B denote the event that the heat absorbed is

above target.

Determine the following probabilities.

a) P(A ∩ B)

b) P(A′)

c) P(A ∪ B

d) P(A ∪ B′)

e) P(A′ ∩ B′)

Answer:

a) P(A ∩ B) = (40 + 16)/204 = 0.2745

b) P(A′) = (36 + 56)/204 = 0.4510

c) P(A ∪ B) = (40 + 12 + 16 + 44 + 36)/204 = 0.7255

d) P(A ∪ B′) = (40 + 12 + 16 + 44 + 56)/204 = 0.8235

e) P(A′ ∩ B′) = 56/204 = 0.2745

2-71. A Web ad can be designed from four different colors,

three font types, five font sizes, three images, and five text

phrases. A specific design is randomly generated by the Web

server when you visit the site. If you visit the site five times, what

is the probability that you will not see the same design?

Answer:

Total number of possible designs is 900. The sample space of all possible designs that may be seen on five visits. This space contains 9005 outcomes.

The number of outcomes in which all five visits are different can be obtained as follows. On the first visit any one of 900 designs may be seen. On the second visit there are 899 remaining designs. On the third visit there are 898 remaining designs. On the fourth and fifth visits there are 897 and 896 remaining designs, respectively. From the multiplication rule, the number of outcomes where all designs are different is 900*899*898*897*896. Therefore, the probability that a design is not seen again is

(900*899*898*897*896)/ 9005 = 0.9889

2-72. Consider the hospital emergency room data in

Example 2-8. Let A denote the event that a visit is to Hospital

4 and let B denote the event that a visit results in LWBS (at any

hospital). Determine the following probabilities

a) P(A ∩ B)

b) P(A′)

c) P(A ∪ B)

d) P(A ∪ B′)

e) P(A′ ∩ B′)

Answer:

a) P(A ∩ B) = 242/22252 = 0.0109

b) P(A′) = (5292+6991+5640)/22252 = 0.8055

c) P(A ∪ B) = (195 + 270 + 246 + 242 + 984 + 3103)/22252 = 0.2265

d) P(A ∪ B′) = (4329 + (5292 – 195) + (6991 – 270) + 5640 – 246))/22252 = 0.9680

e) P(A′ ∩ B′) = (1277 + 1558 + 666 + 3820 + 5163 + 4728)/22252 = 0.7735

Bab 3

3-70. The probability of an operator entering alphanumeric

data incorrectly into a field in a database is equally

likely. The random variable X is the number of fields on a

data entry form with an error. The data entry form has

28 fields. Is X a discrete uniform random variable? Why or

why not?

Answer:

X is a discrete random variable because it denotes the number of fields out of 28 that are in error.

However, X is not uniform because P(X = 0) ≠ P(X = 1).

3-71. Suppose that X has a discrete uniform distribution on

the integers 0 through 9. Determine the mean, variance, and

standard deviation of the random variable Y = 5X and compare

to the corresponding results for X.

Answer:

The range of Y is 0, 5, 10, ..., 45, E(X) = (0+9)/2 = 4.5

E(Y) = 0(1/10)+5(1/10)+...+45(1/10)

= 5[0(0.1) +1(0.1)+ ... +9(0.1)]

= 5E(X)

= 5(4.5)

= 22.5

V(X) = 8.25, V(Y) = 5²(8.25) = 206.25, σ_Y = 14.36

3-72. Show that for a discrete uniform random variable X,

if each of the values in the range of X is multiplied by the

constant c, the effect is to multiply the mean of X by c and

the variance of X by c² . That is, show that E(cX)=cE(X)

and V(cX)= c²V(X) .

Answer:

Bab 4

4-70. The diameter of the dot produced by a printer is normally

distributed with a mean diameter of 0.002 inch and a

standard deviation of 0.0004 inch.

(a) What is the probability that the diameter of a dot exceeds

0.0026 inch?

(b) What is the probability that a diameter is between 0.0014

and 0.0026 inch?

(c) What standard deviation of diameters is needed so that the

probability in part (b) is 0.995?

Answer:

4-71. The weight of a sophisticated running shoe is normally

distributed with a mean of 12 ounces and a standard

deviation of 0.5 ounce.

(a) What is the probability that a shoe weighs more than 13

ounces?

(b) What must the standard deviation of weight be in order for

the company to state that 99.9% of its shoes are less than

13 ounces?

(c) If the standard deviation remains at 0.5 ounce, what must

the mean weight be in order for the company to state that

99.9% of its shoes are less than 13 ounces?

Answer:

4-72. Measurement error that is normally distributed with a

mean of zero and a standard deviation of 0.5 gram is added to

the true weight of a sample. Then the measurement is rounded

to the nearest gram. Suppose that the true weight of a sample

is 165.5 grams.

(a) What is the probability that the rounded result is 167 grams?

(b) What is the probability that the rounded result is 167

grams or greater?

Answer:

Bab 5

5-70. Suppose that X has a uniform probability distribution

Show that the probability distribution of the random variable

Y=2 ln X is chi-squared with two degrees of freedom.

Answer:

5-71. A random variable X has the following probability

distribution:

(a) Find the probability distribution for = X².

(b) Find the probability distribution for Y = X^1/2 .

(c) Find the probability distribution for Y =ln X.

Answer:

5-72. The velocity of a particle in a gas is a random variable

V with probability distribution

where b is a constant that depends on the temperature of the

gas and the mass of the particle.

(a) Find the value of the constant a.

(b) The kinetic energy of the particle is W=mv²/2 . Find the

probability distribution of W.

Answer: