Question 1

Suppose the likelihood that someone who logs onto a particular site on the Internet will purchase an item is 0.10.If the site has 200 people accessing it in the next day,what is the probability that less than 24 customers will make a purchase?
A)0.5008
B)0.1736
C)0.8264
D)0.2461

Accepted Answer

This is a binomial probability problem where n = 200, p = 0.10 (probability of success), and we are looking for P(X < 24). Using a binomial distribution calculator or approximation methods (like the normal approximation to the binomial distribution), we find that the probability of fewer than 24 customers making a purchase is approximately 0.8264.

Question 2

Checkout times in a grocery store follow the exponential distribution with a mean time of 6.5 minutes.What is the probability that the checkout time (T)for the next customer will be less than 5 minutes?
A)0.537
B)0.463
C)0.257
D)0.950

Accepted Answer

The exponential distribution is given by the formula $P(T < t) = 1 - e^{-\lambda t}$, where $\lambda = \frac{1}{\text{mean}}$. Here, the mean checkout time is 6.5 minutes, so $\lambda = \frac{1}{6.5}$. For $t = 5$ minutes, $P(T < 5) = 1 - e^{-\frac{5}{6.5}} \approx 0.537$.

Question 3

A very large logging operation has serious problems keeping their skidders operating properly.The equipment fails at the rate of 3 breakdowns every 48 hours.Assume that x is time between breakdowns and is exponentially distributed.The probability of a single breakdown within 24 hours is:
A)0.0498
B)0.7769
C)0.2231
D)0.9502

Accepted Answer

The exponential distribution is given by the formula $P(X \leq x) = 1 - e^{-\lambda x}$, where $\lambda$ is the rate of occurrence. Given 3 breakdowns every 48 hours, the rate $\lambda = 3 / 48 = 1/16$ breakdowns per hour. For a 24-hour period, $x = 24$, so $P(X \leq 24) = 1 - e^{-(1/16) * 24} = 1 - e^{-1.5} \approx 0.7769$.

Question 4

If Z is a standard normal random variable,the area between Z = 0.0 and Z = 1.25 compared to the area between Z = 1.25 and Z = 2.5 will be:
A)smaller.
B)larger.
C)the same.
D)There is not enough information to answer this question.

Accepted Answer

As we move towards the right on the standard normal distribution curve, the area under the curve decreases. Therefore, the area between Z = 1.25 and Z = 2.5 is smaller than the area between Z = 0.0 and Z = 1.25, and the answer is B: larger.

Question 5

Which of the following is not true for an exponential distribution with parameter λ?
A)Mean = 1/λ
B)Standard deviation = 1/λ
C)The distribution is completely determined once the value of λ is known.
D)The distribution is symmetric around the mean.

Accepted Answer

The exponential distribution is not symmetric around the mean. In fact, it is skewed to the right. The mean and standard deviation are both equal to 1/λ, and the distribution is completely determined once the value of λ is known.

Question 6

Let the random variable Z follow a standard normal distribution,and let z₁be a possible value of Z that is unknown but identifiable by position and area.Find z₁if the area to the right of z₁ is 0.0869. A)-1.36 B)1.71 C)1.36 D)1.80

Accepted Answer

Since the area to the right of z is 0.0869, the area to the left of z is 1 – 0.0869 = 0.9131. Using a standard normal distribution table or calculator, we can find the z-score that corresponds to an area of 0.9131 to be approximately 1.36. Therefore, z₁ = 1.36.

Question 7

If the mean of an exponential distribution is 2,then the value of the parameter λ is:
A)4.0
B)2.2
C)1.0
D)0.5

Accepted Answer

The mean ($\mu$) of an exponential distribution is given by $1/\lambda$. Given that the mean is 2, we solve $1/\lambda = 2$ to find $\lambda = 0.5$.

Question 8

If Z is a standard normal random variable,the area between Z = 0.0 and Z =1.20 is 0.3849,while the area between Z = 0.0 and Z = 1.40 is 0.4192.What is the area between Z = -1.20 and Z = 1.40?
A)0.8041
B)0.0808
C)0.1151
D)0.0343

Accepted Answer

The area between Z = -1.20 and Z = 1.40 is found by adding the area from Z = 0 to Z = 1.20 (0.3849) and the area from Z = 0 to Z = 1.40 (0.4192), then subtracting the overlap (the area from Z = 0 to Z = 1.20, since it's included in both calculations). However, since the normal distribution is symmetric, the area from Z = 0 to Z = -1.20 is the same as from Z = 0 to Z = 1.20. Therefore, we simply add the area from Z = 0 to Z = 1.20 (0.3849) to the area from Z = 0 to Z = 1.40 (0.4192) to get the total area, which equals 0.8041.

Question 9

If X is a normal random variable with mean of 1228 and a standard deviation of 120,the number of standard deviations from 1228 to 1380 is:
A)10.233
B)3.1989
C)11.50
D)1.267

Accepted Answer

The number of standard deviations from the mean to a given value is calculated by subtracting the mean from the value and then dividing by the standard deviation. For this case, (1380 - 1228) / 120 = 152 / 120 = 1.267.

Question 10

Let the random variable Z follow a standard normal distribution,and let z₁ be a possible value of Z that is unknown but identifiable by position and area.Find z₁ if the symmetrical area between a negative z₁ and a positive z₁ is 0.9544. A)2.00 B)0.11 C)2.50 D)0.06

Accepted Answer

Since the area between a negative z-score and a positive z-score is 0.9544 and the normal distribution is symmetrical, we can find the z-score corresponding to the area (0.9544 + 1)/2 = 0.9772 using a standard normal distribution table or a calculator. Using a standard normal distribution table, we find that the z-score corresponding to an area of 0.9772 is approximately 2.00 (rounding to two decimal places). Therefore, z₁ = 2.00. Alternatively, using a calculator, we can use the inverse normal function to find the z-score corresponding to an area of 0.9772. In Excel, the function is =NORMSINV(0.9772), which also gives a result of approximately 2.00.

Question 11

Given that Z is a standard normal random variable,P(-1.2 ≤ Z ≤ 1.5)is:
A)0.8181
B)0.4772
C)0.3849
D)0.5228

Accepted Answer

To find $P(-1.2 \leq Z \leq 1.5)$, we look up the Z-scores in the standard normal distribution table or use a calculator. The value for $Z = 1.5$ is approximately 0.9332, and for $Z = -1.2$, it is approximately 0.1151. Subtracting these, $0.9332 - 0.1151 = 0.8181$, gives the probability of Z falling between -1.2 and 1.5.

Question 12

Let the random variable Z follow a standard normal distribution,and let z₁be a possible value of Z that is representing the 90th percentile of the standard normal distribution.Find the value of z_1. A)0.67 B)-1.28 C)1.28 D)-0.67

Accepted Answer

By definition, the 90th percentile of a standard normal distribution means that 90% of the area under the curve lies below the corresponding z-value. Using a standard normal probability table or calculator, we can find that the z-value corresponding to a cumulative area of 0.9 is 1.28. Therefore, the value of z corresponding to the 90th percentile is 1.28.

Question 13

If the random variable X is exponentially distributed with parameter λ = 4,then the probability P(X ≤ 0.25)is equal to:
A)0.6321
B)0.3679
C)0.2500
D)0.5000

Accepted Answer

The cumulative distribution function (CDF) for an exponential distribution with parameter λ is given by F(x) = 1 - e^(-λx). 
Thus, for λ = 4, we have F(x) = 1 - e^(-4x). 
To find P(X ≤ 0.25), we simply evaluate F(0.25): 
P(X ≤ 0.25) = F(0.25) = 1 - e^(-4*0.25) = 1 - e^(-1) = 0.6321.

Therefore, the answer is choice A.

Question 14

Let the random variable Z follow a standard normal distribution,and let z₁be a possible value of Z that is representing the 75th percentile of the standard normal distribution.Find the value of z_1. A)0.67 B)-0.67 C)1.28 D)-1.28

Accepted Answer

The 75th percentile of the standard normal distribution corresponds to a z-score of 0.67.

Question 15

The owner of a fish market determined that the average weight for a catfish is 3.4 pounds with a standard deviation of 0.8 of a pound.A citation catfish should be one of the top 2% in weight.Assuming the weights of catfish are normally distributed,at what weight (in pounds)should the citation designation be established?
A)1.65 pounds
B)5.04 pounds
C)5.42 pounds
D)6.76 pounds

Accepted Answer

To find the weight at which a catfish would be in the top 2% (or the 98th percentile) in a normal distribution, we use the Z-score formula: Z = (X - μ) / σ, where X is the value in the distribution, μ is the mean, and σ is the standard deviation. For the top 2%, the Z-score is approximately 2.05 (from standard Z-tables). Plugging in the values: 2.05 = (X - 3.4) / 0.8, solving for X gives X ≈ 5.04 pounds.

Question 16

Let the random variable Z follow a standard normal distribution,and let z₁ be a possible value of Z that is representing the 10th percentile of the standard normal distribution.Find the value of z_1. A)0.255 B)-0.255 C)1.28 D)-1.28

Accepted Answer

The 10th percentile of a standard normal distribution corresponds to a z-score of approximately -1.28.

Question 17

If the random variable x is exponentially distributed with parameter λ = 1.5,then the probability P(2 ≤ X ≤ 4)is equal to:
A)0.6667
B)0.5000
C)0.0473
D)0.2500

Accepted Answer

The probability density function of an exponential distribution with parameter λ is given by:

f(x) = λe^(-λx), for x ≥ 0

So, for λ = 1.5, we have:

f(x) = 1.5e^(-1.5x), for x ≥ 0

The cumulative distribution function is:

F(x) = ∫₀ˣ f(t) dt = 1 - e^(-λx)

So, for λ = 1.5, we have:

F(x) = 1 - e^(-1.5x)

Now, to calculate P(2 ≤ X ≤ 4), we need to subtract the cumulative probabilities at x=2 from the cumulative probability at x=4:

P(2 ≤ X ≤ 4) = F(4) - F(2)
             = (1 - e^(-6)) - (1 - e^(-3))
             = e^(-3) - e^(-6)
             ≈ 0.0473

Therefore, the answer is C.

Question 18

Which of the following is not a continuous distribution?
A)normal distribution
B)exponential distribution
C)uniform distribution
D)Poisson distribution

Accepted Answer

The Poisson distribution is a discrete distribution, as it models the number of occurrences of an event in a fixed unit of time or space, which can only take on integer values. The other options are continuous distributions, as they model continuous random variables that can take on any value within a given range.

Question 19

If the continuous random variable X is uniformly distributed over the interval 6 to 10,then the height of this distribution,f(x),is:
A)0.125
B)0.250
C)0.500
D)0.750

Accepted Answer

For a uniform distribution over an interval [a, b], the height of the distribution, f(x), is given by 1/(b-a). Here, a = 6 and b = 10, so f(x) = 1/(10-6) = 1/4 = 0.250.

Question 20

A catalog company administered a survey to determine how long customers were willing to wait on hold before ordering a product.The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3.5 minutes.What proportion of customers having to hold more than 5.0 minutes will hang up before placing an order?
A)0.2397
B)0.5714
C)0.4286
D)0.7603

Accepted Answer

Let X be the length of time a customer waits on hold. We know that X follows an exponential distribution with mean 3.5 minutes, i.e. E(X) = 3.5.

We need to find P(X > 5 | X > 5), i.e. the probability that a customer hangs up given that they have already waited more than 5 minutes.

Using the memoryless property of the exponential distribution, we have:

P(X > 5 | X > 5) = P(X > 5)/P(X > 5) = e^(-5/3.5) / e^(-5/3.5) = e^(-5/3.5) ≈ 0.2397.

Therefore, the proportion of customers hanging up after waiting more than 5.0 minutes is approximately 0.2397, which corresponds to choice A.

Quiz 5: Continuous Random Variables and Probability Distributions

Suppose the likelihood that someone who logs onto a particular site on the Internet will purchase an item is 0.10.If the site has 200 people accessing it in the next day,what is the probability that less than 24 customers will make a purchase?

Checkout times in a grocery store follow the exponential distribution with a mean time of 6.5 minutes.What is the probability that the checkout time (T) for the next customer will be less than 5 minutes?

A very large logging operation has serious problems keeping their skidders operating properly.The equipment fails at the rate of 3 breakdowns every 48 hours.Assume that x is time between breakdowns and is exponentially distributed.The probability of a single breakdown within 24 hours is:

If Z is a standard normal random variable,the area between Z = 0.0 and Z = 1.25 compared to the area between Z = 1.25 and Z = 2.5 will be:

Which of the following is not true for an exponential distribution with parameter λ?

Let the random variable Z follow a standard normal distribution,and let z₁be a possible value of Z that is unknown but identifiable by position and area.Find z₁if the area to the right of z₁ is 0.0869.

If the mean of an exponential distribution is 2,then the value of the parameter λ is:

If Z is a standard normal random variable,the area between Z = 0.0 and Z =1.20 is 0.3849,while the area between Z = 0.0 and Z = 1.40 is 0.4192.What is the area between Z = -1.20 and Z = 1.40?

If X is a normal random variable with mean of 1228 and a standard deviation of 120,the number of standard deviations from 1228 to 1380 is:

Let the random variable Z follow a standard normal distribution,and let z₁ be a possible value of Z that is unknown but identifiable by position and area.Find z₁ if the symmetrical area between a negative z₁ and a positive z₁ is 0.9544.

Given that Z is a standard normal random variable,P(-1.2 ≤ Z ≤ 1.5) is:

Let the random variable Z follow a standard normal distribution,and let z₁be a possible value of Z that is representing the 90th percentile of the standard normal distribution.Find the value of z_1.

If the random variable X is exponentially distributed with parameter λ = 4,then the probability P(X ≤ 0.25) is equal to:

Let the random variable Z follow a standard normal distribution,and let z₁be a possible value of Z that is representing the 75th percentile of the standard normal distribution.Find the value of z_1.

Let the random variable Z follow a standard normal distribution,and let z₁ be a possible value of Z that is representing the 10th percentile of the standard normal distribution.Find the value of z_1.

If the random variable x is exponentially distributed with parameter λ = 1.5,then the probability P(2 ≤ X ≤ 4) is equal to:

Which of the following is not a continuous distribution?

If the continuous random variable X is uniformly distributed over the interval 6 to 10,then the height of this distribution,f(x) ,is: