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Solved Problems

Solved Problem 2-1 In the past 30 days, Roger’s Rural Roundup has sold 8, 9, 10, or 11 lottery tickets. It never sold fewer than 8 or more than 11. Assuming that the past is similar to the future, find the probabilities for the number of tickets sold if sales were 8 tickets on 10 days, 9 tickets on 12 days, 10 tickets on 6 days, and 11 tickets on 2 days.

Solution

SALES NO. DAYS PROBABILITY

8 10 0.333

9 12 0.400

10 6 0.200

11 2 0.067

Total 30 1.000
Solved Problem 2-2 A class contains 30 students. Ten are female (F) and U.S. citizens (U); 12 are male (M) and U.S. citizens; 6 are female and non-U.S. citizens (N); 2 are male and non-U.S. citizens.

A name is randomly selected from the class roster, and it is female. What is the probability that the student is a U.S. citizen?

Solution

$\begin{array}{l} P (FU) & = & \frac{10}{30} = 0.333 \\ P (FN) & = & \frac{6}{30} = 0.200 \\ P (MU) & = & \frac{12}{30} = 0.400 \\ P (MN) & = & \frac{2}{30} = 0.067 \\ P (F) & = & P (FU) + P (FN) = 0.333 + 0.200 = 0.533 \\ P (M) & = & P (MU) + P (MN) = 0.400 + 0.067 = 0.467 \\ P (U) & = & P (FU) + P (MU) = 0.333 + 0.400 = 0.733 \\ P (N) & = & P (FN) + P (MN) = 0.200 + 0.067 = 0.267 \\ P (U | F) & = & \frac{P (FU)}{P (F)} = \frac{0.333}{0.533} = 0.625 \end{array}$ $\begin{array}{l} P (FU) & = & \frac{10}{30} = 0.333 \\ P (FN) & = & \frac{6}{30} = 0.200 \\ P (MU) & = & \frac{12}{30} = 0.400 \\ P (MN) & = & \frac{2}{30} = 0.067 \\ P (F) & = & P (FU) + P (FN) = 0.333 + 0.200 = 0.533 \\ P (M) & = & P (MU) + P (MN) = 0.400 + 0.067 = 0.467 \\ P (U) & = & P (FU) + P (MU) = 0.333 + 0.400 = 0.733 \\ P (N) & = & P (FN) + P (MN) = 0.200 + 0.067 = 0.267 \\ P (U | F) & = & \frac{P (FU)}{P (F)} = \frac{0.333}{0.533} = 0.625 \end{array}$
Solved Problem 2-3 Your professor tells you that if you score an 85 or better on your midterm exam, then you have a 90% chance of getting an A for the course. You think you have only a 50% chance of scoring 85 or better. Find the probability that both your score is 85 or better and you receive an A in the course.

Solution

$\begin{array}{l} P (A and 85) & = & P (A | 85) \times P (85) = (0.90) (0.50) \\ = & 40 % \end{array}$ $\begin{array}{l} P (A and 85) & = & P (A | 85) \times P (85) = (0.90) (0.50) \\ = & 40 % \end{array}$

SALES	NO. DAYS	PROBABILITY
8	10	0.333
9	12	0.400
10	6	0.200
11	2	0.067
Total	30	1.000

Solved Problem 2-4 A statistics class was asked if it believed that all tests on the Monday following the football game win over their archrival should be postponed automatically. The results were as follows:

Strongly agree	40
Agree	30
Neutral	20
Disagree	10
Strongly disagree	0
	100

Transform this into a numeric score, using the following random variable scale, and find a probability distribution for the results:

Strongly agree		5
Agree		4
Neutral		3
Disagree		2
Strongly disagree		1

Solution

OUTCOME	PROBABILITY, P (X)
Strongly agree (5)	0.4 $=$ $=$ 40 $/$ $/$ 100
Agree (4)	0.3 $=$ $=$ 30 $/$ $/$ 100
Neutral (3)	0.2 $=$ $=$ 20 $/$ $/$ 100
Disagree (2)	0.1 $=$ $=$ 10 $/$ $/$ 100
Strongly disagree (1)	0.0 $=$ $=$ 0 $/$ $/$ 100
Total	1.0 $=$ $=$ 100 $/$ $/$ 100

Solved Problem 2-5 For Solved Problem 2-4, let X be the numeric score. Compute the expected value of X.

Solution

$\begin{array}{l} E (X) & = & \sum_{i = 1}^{5} X_{i} P (X_{i}) = X_{1} P (X_{1}) + X_{2} P (X_{2}) \\ + X_{3} P (X_{3}) + X_{4} P (X_{4}) + X_{5} P (X_{5}) \\ = & 5 (0.4) + 4 (0.3) + 3 (0.2) + 2 (0.1) + 1 (0) \\ = & 4.0 \end{array}$ $\begin{array}{l} E (X) & = & \sum_{i = 1}^{5} X_{i} P (X_{i}) = X_{1} P (X_{1}) + X_{2} P (X_{2}) \\ + X_{3} P (X_{3}) + X_{4} P (X_{4}) + X_{5} P (X_{5}) \\ = & 5 (0.4) + 4 (0.3) + 3 (0.2) + 2 (0.1) + 1 (0) \\ = & 4.0 \end{array}$
Solved Problem 2-6 Compute the variance and standard deviation for the random variable X in Solved Problems 2-4 and 2-5.

Solution

$\begin{array}{l} Variance & = & \sum_{i = 1}^{5} {(X_{i} - E (X))}^{2} P (X_{i}) \\ = & {(5 - 4)}^{2} (0.4) + {(4 - 4)}^{2} (0.3) + {(3 - 4)}^{2} (0.2) + {(2 - 4)}^{2} (0.1) + {(1 - 4)}^{2} (0.0) \\ = & {(1)}^{2} (0.4) + {(0)}^{2} (0.3) + {(- 1)}^{2} (0.2) + {(- 2)}^{2} (0.1) + {(- 3)}^{2} (0.0) \\ = & 0.4 + 0.0 + 0.2 + 0.4 + 0.0 = 1.0 \end{array}$ $\begin{array}{l} Variance & = & \sum_{i = 1}^{5} {(X_{i} - E (X))}^{2} P (X_{i}) \\ = & {(5 - 4)}^{2} (0.4) + {(4 - 4)}^{2} (0.3) + {(3 - 4)}^{2} (0.2) + {(2 - 4)}^{2} (0.1) + {(1 - 4)}^{2} (0.0) \\ = & {(1)}^{2} (0.4) + {(0)}^{2} (0.3) + {(- 1)}^{2} (0.2) + {(- 2)}^{2} (0.1) + {(- 3)}^{2} (0.0) \\ = & 0.4 + 0.0 + 0.2 + 0.4 + 0.0 = 1.0 \end{array}$

The standard deviation is

$σ = \sqrt{Variance} = \sqrt{1} = 1$ $σ = \sqrt{Variance} = \sqrt{1} = 1$
Solved Problem 2-7 A candidate for public office has claimed that 60% of voters will vote for her. If 5 registered voters were sampled, what is the probability that exactly 3 would say they favor this candidate?

Solution

We use the binomial distribution with $n = 5, p = 0.6,$ $n = 5, p = 0.6,$ and $r = 3 :$ $r = 3 :$

$P (exactly 3 successes in 5 trials) = \frac{n!}{r! (n - r)!} p^{r} q^{n - r} = \frac{5!}{3! (5 - 3)!} {(0.6)}^{3} {(0.4)}^{5 - 3} = 0.3456$ $P (exactly 3 successes in 5 trials) = \frac{n!}{r! (n - r)!} p^{r} q^{n - r} = \frac{5!}{3! (5 - 3)!} {(0.6)}^{3} {(0.4)}^{5 - 3} = 0.3456$
Solved Problem 2-8 The length of the rods coming out of our new cutting machine can be said to approximate a normal distribution with a mean of 10 inches and a standard deviation of 0.2 inch. Find the probability that a rod selected randomly will have a length
1. of less than 10.0 inches.
2. between 10.0 and 10.4 inches.
3. between 10.0 and 10.1 inches.
4. between 10.1 and 10.4 inches.
5. between 9.6 and 9.9 inches.
6. between 9.9 and 10.4 inches.
7. between 9.886 and 10.406 inches.
Solution

First, compute the standard normal distribution, the Z value:

$Z = \frac{X - μ}{σ}$ $Z = \frac{X - μ}{σ}$

Next, find the area under the curve for the given Z value by using a standard normal distribution table.
1. $P (X < 10.0) = 0.50000$ $P (X < 10.0) = 0.50000$
2. $P (10.0 < X < 10.4) = 0.97725 - 0.50000 = 0.47725$ $P (10.0 < X < 10.4) = 0.97725 - 0.50000 = 0.47725$
3. $P (10.0 < X < 10.1) = 0.69146 - 0.50000 = 0.19146$ $P (10.0 < X < 10.1) = 0.69146 - 0.50000 = 0.19146$
4. $P (10.1 < X < 10.4) = 0.97725 - 0.69146 = 0.28579$ $P (10.1 < X < 10.4) = 0.97725 - 0.69146 = 0.28579$
5. $P (9.6 < X < 9.9) = 0.97725 - 0.69146 = 0.28579$ $P (9.6 < X < 9.9) = 0.97725 - 0.69146 = 0.28579$
6. $P (9.9 < X < 10.4) = 0.19146 + 0.47725 = 0.66871$ $P (9.9 < X < 10.4) = 0.19146 + 0.47725 = 0.66871$
7. $P (9.886 < X < 10.406) = 0.47882 + 0.21566 = 0.69448$ $P (9.886 < X < 10.406) = 0.47882 + 0.21566 = 0.69448$

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