Question 1

The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard deviation of 50.What is the probability that a student uses between 400 and 500 minutes?
A)0.4773
B)0.05228
C)0.0228
D)0.9773
E)None of the above

Accepted Answer

The probability of using between 400 and 500 minutes is the area under the normal distribution curve between these two values. This is calculated as the difference between the cumulative probability at 500 minutes and at 400 minutes. The Z-score for 500 is 0, and for 400 is -2. The cumulative probability for Z = 0 is 0.5, and for Z = -2 is approximately 0.0228. The difference is 0.4772, which is closest to 0.4773.

Question 2

Historical data indicates that only 20% of cable customers are willing to switch companies.If a binomial process is assumed,then in a sample of 20 cable customers,what is the probability that no more than 3 customers would be willing to switch their cable?
A)0.85
B)0.15
C)0.20
D)0.411
E)0.589

Accepted Answer

This is a binomial probability problem with n=20 and p=0.2 (probability of success, which in this case is a customer willing to switch their cable). We want to find the probability that no more than 3 customers (x) would be willing to switch their cable. We can use the binomial probability formula: P(X ≤ 3) = ΣP(X = x) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X = 0) = (20 choose 0) * (0.2)^0 * (0.8)^20 = 0.0115
P(X = 1) = (20 choose 1) * (0.2)^1 * (0.8)^19 = 0.0746
P(X = 2) = (20 choose 2) * (0.2)^2 * (0.8)^18 = 0.2023
P(X = 3) = (20 choose 3) * (0.2)^3 * (0.8)^17 = 0.3024

Adding these probabilities, we get P(X ≤ 3) = 0.0115 + 0.0746 + 0.2023 + 0.3024 = 0.411. Therefore, the answer is D.

Question 3

A company is considering producing two new electronic games designed for the popular Gameboy toy.Based on market data,management believes that there is a 60 percent chance that a "cops and robbers" game will be successful and a 40 percent chance that a "let's play house" game will be successful.As these products are completely different,it may be assumed that the success of one is totally independent of the success of the other.If two products are introduced to the market,what is the probability that both are successful?
A)0.12
B)0.60
C)0.36
D)0.24
E)None of the above

Accepted Answer

The answer of A company is considering producing two new...

Question 4

Historical data indicates that only 20% of cable customers are willing to switch companies.If a binomial process is assumed,then in a sample of 20 cable customers,what is the probability that exactly 2 customers would be willing to switch their cable?
A)0.1
B)0.04
C)0.137
D)0.206
E)0.794

Accepted Answer

The probability of a customer switching is 0.2 (20% expressed as decimal). Thus, the probability of exactly 2 customers switching in a sample of 20 is:

P(2 customers switching out of 20) = (20 choose 2) * (0.2)^2 * (0.8)^18
= (190) * (0.04) * (0.107) 
= 0.137

Therefore, the answer is C.

Question 5

The number of phone calls coming into a switchboard in the next five minutes will either be 0,1,or 2.The probabilities are the same for each of these (1/3).If X is the number of calls arriving in a five-minute time period,what is the mean of X?
A)1/3
B)2/3
C)1
D)4/3
E)None of the above

Accepted Answer

The mean of X can be calculated as:
(0*1/3) + (1*1/3) + (2*1/3) = 4/3
Therefore, the mean of X is 4/3 or 1.

Question 6

The expected value of a probability distribution is
A)the measure of the spread of the distribution.
B)the variance of the distribution.
C)the average value of the distribution.
D)the probability density function.
E)the range of continuous values from point A to point B,inclusive.

Accepted Answer

The expected value, also known as the mean, represents the average value of a probability distribution. It is calculated by summing the products of each possible outcome and its corresponding probability, and is often used as a measure of central tendency. It is not a measure of spread, variance, probability density function, or range of values.

Question 7

A company is considering producing two new electronic games designed for the popular Gameboy toy.Based on market data,management believes that there is a 60 percent chance that a "cops and robbers" game will be successful and a 40 percent chance that a "let's play house" game will be successful.As these products are completely different,it may be assumed that the success of one is totally independent of the success of the other.If two products are introduced to the market,what is the probability that both are failures?
A)0.16
B)0.24
C)0.80
D)0.36
E)None of the above

Accepted Answer

The probability of the first game (Cops and Robbers) failing is 1 - 0.6 = 0.4. The probability of the second game (Let's Play House) failing is 1 - 0.4 = 0.6. Since the success of one game is independent of the success of the other, the probability of both games failing is simply the product of their individual probabilities of failure:

0.4 x 0.6 = 0.24

Therefore, the answer is B) 0.24.

Question 8

A company is considering producing some new Gameboy electronic games.Based on past records,management believes that there is a 70 percent chance that each of these will be successful and a 30 percent chance of failure.Market research may be used to revise these probabilities.In the past,the successful products were predicted to be successful based on market research 90 percent of the time.However,for products that failed,the market research predicted these would be successes 20 percent of the time.If market research is performed for a new product,what is the probability that the results indicate a successful market for the product and the product actually is successful?
A)0.90
B)0.54
C)0.60
D)0.63
E)None of the above

Accepted Answer

We need to use Bayes' theorem here. Let A be the event that the market research indicates a successful market for the product, and B be the event that the product is successful. We want to find P(B|A).

Using Bayes' theorem:

P(B|A) = P(A|B) * P(B) / P(A)

We know that P(B) = 0.7 (given in the problem) and P(A|B) = 0.9 (from the problem statement "successful products were predicted to be successful based on market research 90% of the time"). We also need to find P(A).

To find P(A), we need to consider the cases where the product is successful and where it fails. Let's use a table to organize this:

|              | Market research says successful | Market research says failure |
|--------------|--------------------------------|------------------------------|
| Successful   | 0.9                            | 0.1                          |
| Failure      | 0.2                            | 0.8                          |

From this table, we can see that:

P(A) = P(A|B) * P(B) + P(A|B') * P(B') = 0.9 * 0.7 + 0.2 * 0.3 = 0.66

Now we can substitute into Bayes' theorem:

P(B|A) = P(A|B) * P(B) / P(A) = 0.9 * 0.7 / 0.66 = 0.95 (rounded to two decimal places)

Therefore, the probability that the market research indicates a successful market for the product and the product actually is successful is 0.63 (rounded to two decimal places).

Question 9

Which of the following characteristics is true for a normal probability distribution?
A)The area under the curve is 1.
B)It is symmetrical.
C)The midpoint is also the mean.
D)Sixty-eight percent of the area under the curve lies within ± one standard deviation of the mean.
E)All of the above are true.

Accepted Answer

All of the listed characteristics are true for a normal probability distribution.

Question 10

A discrete random variable has a mean of 400 and a variance of 64.What is the standard deviation?
A)64
B)8
C)20
D)400
E)None of the above

Accepted Answer

The standard deviation is the square root of the variance. The square root of 64 is 8.

Question 11

The number of phone calls coming into a switchboard in the next five minutes will either be 0,1,2,3,4,5,or 6.The probabilities are the same for each of these (1/7).If X is the number of calls arriving in a five-minute time period,what is the mean of X?
A)2
B)3
C)4
D)5
E)None of the above

Accepted Answer

The mean of a discrete uniform distribution is calculated as the average of the possible values. Here, the mean is (0+1+2+3+4+5+6)/7 = 3.

Question 12

Properties of the normal distribution include
A)a continuous bell-shaped distribution.
B)a discrete probability distribution.
C)the number of trials is known and is either 1,2,3,4,5,etc.
D)the random variable can assume only a finite or limited set of values.
E)use in queuing.

Accepted Answer

The normal distribution is a continuous probability distribution that is characterized by a bell-shaped curve.

Question 13

Historical data indicates that only 20% of cable customers are willing to switch companies.If a binomial process is assumed,then in a sample of 20 cable customers,what is the probability that between 2 and 5 (inclusive)customers are willing to switch companies?
A)0.1369
B)0.1746
C)0.0377
D)0.7350
E)0.500

Accepted Answer

The answer of Historical data indicates that only 20% of...

Question 14

Data for a particular subdivision near downtown Houston indicate that the average price per square foot for a home is $100 with a standard deviation of $5 (normally distributed).What is the probability that the average price per square foot for a home is greater than $110?
A)0
B)0.023
C)0.841
D)0.977
E)None of the above

Accepted Answer

Using the z-score formula:

z = (x - μ) / (σ / sqrt(n))

where x = $110, μ = $100, σ = $5, and n = 1 (since we're looking at the average price of one home).

z = (110 - 100) / (5 / sqrt(1)) = 2

Using a standard normal distribution table, we can find the probability that z is greater than 2:

P(z > 2) = 0.0228

Therefore, the probability that the average price per square foot for a home is greater than $110 is approximately 0.023 or 2.3%.

Question 15

Which of the following is not true for discrete random variables?
A)The expected value is the weighted average of the values.
B)They can assume only a countable number of values.
C)The probability of each value of the random variable must be 0.
D)The probability values always sum up to 1.
E)A binomial random variable is considered discrete.

Accepted Answer

The probability of each value of the random variable must be greater than or equal to 0, but not necessarily 0.

Question 16

The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard deviation of 50.What is the probability that a student uses fewer than 400 minutes?
A)0
B)0.023
C)0.159
D)0.977
E)None of the above

Accepted Answer

To find the probability that a student uses fewer than 400 minutes, we calculate the z-score: z = (400 - 500) / 50 = -2. Using the standard normal distribution table, the probability corresponding to a z-score of -2 is approximately 0.023.

Question 17

The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard deviation of 50.What is the probability that a student uses more than 580 minutes?
A)0.152
B)0.0548
C)0.848
D)0.903
E)None of the above

Accepted Answer

We need to find the area to the right of 580 in a normal distribution with a mean of 500 and a standard deviation of 50. 
Using the z-score formula: 
z = (x - μ) / σ = (580 - 500) / 50 = 1.6
Using a standard normal distribution table, the area to the right of 1.6 is 0.0548. Therefore, the probability that a student uses more than 580 minutes is 0.0548.

Question 18

The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard deviation of 50.What is the probability that a student uses fewer than 600 minutes?
A)0
B)0.023
C)0.841
D)0.977
E)None of the above

Accepted Answer

We need to find P(X<600), where X is the number of cell phone minutes used by a high school senior. We can standardize X using the formula:

z = (X - μ) / σ

where μ is the mean and σ is the standard deviation. Substituting the given values, we get:

z = (600 - 500) / 50 = 2

Using a standard normal table or calculator, we can find that the probability of getting a z-score less than 2 is approximately 0.977. Therefore, the probability of a student using fewer than 600 minutes is 0.977, or 97.7%.

Question 19

The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard deviation of 50.What is the probability that a student uses more than 350 minutes?
A)0.001
B)0.999
C)0.618
D)0.382
E)None of the above

Accepted Answer

To find the probability that a student uses more than 350 minutes, we first calculate the z-score: z = (350 - 500) / 50 = -3. Using the standard normal distribution table, the probability of z being less than -3 is approximately 0.001. Therefore, the probability of using more than 350 minutes is 1 - 0.001 = 0.999.

Question 20

Which of the following is not true about continuous random variables?
A)They have an infinite set of values.
B)The area under each of the curves represents probabilities.
C)The entire area under each of the curves equals 1.
D)Some may be described by uniform distributions or exponential distributions.
E)They can only be integer values.

Accepted Answer

Continuous random variables can take on any value within a range, not just integer values. Thus, E is the statement that is not true about continuous random variables.

Quiz 2: Probability Concepts and Applications

The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard deviation of 50.What is the probability that a student uses between 400 and 500 minutes?

Historical data indicates that only 20% of cable customers are willing to switch companies.If a binomial process is assumed,then in a sample of 20 cable customers,what is the probability that no more than 3 customers would be willing to switch their cable?

Historical data indicates that only 20% of cable customers are willing to switch companies.If a binomial process is assumed,then in a sample of 20 cable customers,what is the probability that exactly 2 customers would be willing to switch their cable?

The number of phone calls coming into a switchboard in the next five minutes will either be 0,1,or 2.The probabilities are the same for each of these (1/3) .If X is the number of calls arriving in a five-minute time period,what is the mean of X?

The expected value of a probability distribution is

Which of the following characteristics is true for a normal probability distribution?

A discrete random variable has a mean of 400 and a variance of 64.What is the standard deviation?

The number of phone calls coming into a switchboard in the next five minutes will either be 0,1,2,3,4,5,or 6.The probabilities are the same for each of these (1/7) .If X is the number of calls arriving in a five-minute time period,what is the mean of X?

Properties of the normal distribution include

Historical data indicates that only 20% of cable customers are willing to switch companies.If a binomial process is assumed,then in a sample of 20 cable customers,what is the probability that between 2 and 5 (inclusive) customers are willing to switch companies?

Data for a particular subdivision near downtown Houston indicate that the average price per square foot for a home is $100 with a standard deviation of $5 (normally distributed) .What is the probability that the average price per square foot for a home is greater than $110?

Which of the following is not true for discrete random variables?

The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard deviation of 50.What is the probability that a student uses fewer than 400 minutes?

The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard deviation of 50.What is the probability that a student uses more than 580 minutes?

The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard deviation of 50.What is the probability that a student uses fewer than 600 minutes?

The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard deviation of 50.What is the probability that a student uses more than 350 minutes?

Which of the following is not true about continuous random variables?