X is a continuous random variable with density f(x). What is P(X = 5)?
Zero
✓
The density value f(5)
One half
One
Correct Answer
Zero
A single point has no width, so the area above it, and therefore its probability, is 0. The density value f(5) is the height of the curve at 5, not a probability. One half would require a region with real width. One is the total area under the whole curve, not the area at a single point.
Category: Probability as Area
Question 2
For a continuous random variable, what gives P(c < X < d) on the graph of its density function?
The height of the curve at the midpoint of c and d
The area under the curve between the two values
✓
The slope of the curve between the two values
The distance from c to d along the x-axis
Correct Answer
The area under the curve between the two values
For continuous distributions, probability equals area, so P(c < X < d) is the area under f(x) from c to d. The height of the curve at the midpoint of c and d is a density value, not an area. The slope of the curve between the two values says how steeply it rises or falls. The distance from c to d along the x-axis ignores how tall the curve is.
Category: Probability as Area
Question 3
What is the total area under the graph of any probability density function?
0.5
100
1
✓
The mean of X
Correct Answer
1
The total area under a density equals 1 because it represents the probability that X takes some value in its range. An area of 0.5 is only half of the probability, such as the part to the left of the median. A total of 100 confuses probability with percent. The mean of X is a center, and it can be any number, so it is not the total area.
Category: Probability as Area
Question 4
X ~ U(2, 10). What is the height of its probability density function between 2 and 10?
1/10
1/12
8
1/8
✓
Correct Answer
1/8
A uniform density on a to b has height 1/(b - a), here 1/(10 - 2) = 1/8, so the rectangle has area 8 times 1/8 = 1. The value 1/10 wrongly uses only the upper endpoint. The value 1/12 adds the endpoints instead of subtracting them. The value 8 is the width of the rectangle, not its height.
Category: Uniform Calculations
Question 5
X ~ U(2, 10). What is P(4 < X < 7)? Round to three decimal places.
0.375
✓
0.300
0.700
0.875
Correct Answer
0.375
The probability is the area of a rectangle: (7 - 4) times 1/(10 - 2) = 3/8 = 0.375. The value 0.300 divides the width 3 by 10 instead of by the range 8. The value 0.700 divides the upper bound 7 by 10. The value 0.875 is 7/8, which ignores the lower bound of 4.
Category: Uniform Calculations
Question 6
X ~ U(4, 20). What is the mean of X?
8
12
✓
16
10
Correct Answer
12
The mean of a uniform distribution is the midpoint (a + b)/2 = (4 + 20)/2 = 12. The value 8 is (20 - 4)/2, half the width rather than the midpoint. The value 16 is the width b - a. The value 10 is half of the upper endpoint and ignores a.
Category: Uniform Calculations
Question 7
X ~ U(0, 12). What is the standard deviation of X? Round to two decimal places.
12.00
1.00
3.46
✓
6.00
Correct Answer
3.46
The standard deviation is sqrt((b - a)^2/12) = sqrt(144/12) = sqrt(12), which is about 3.46. The value 12.00 is the variance, before taking the square root. The value 1.00 divides the width 12 by 12 without squaring it first. The value 6.00 is half the width, which is the distance from the mean to an endpoint.
Category: Uniform Calculations
Question 8
A random number is drawn uniformly from 0 to 20. Below what value does 75 percent of the distribution lie?
5
10
7.5
15
✓
Correct Answer
15
The kth percentile of a uniform distribution is a + p(b - a) = 0 + 0.75(20) = 15, since 75 percent of the rectangle's area lies to its left. The value 5 is the 25th percentile. The value 10 is the median. The value 7.5 multiplies 0.75 by 10 instead of by the full width of 20.
Category: Uniform Calculations
Question 9
A variable is uniform on 0 to 10. Given that it is greater than 4, what is the chance that it is greater than 6? Give three decimal places.
0.667
✓
0.400
0.600
0.240
Correct Answer
0.667
Given X > 4, X is uniform from 4 to 10, so P(X > 6 | X > 4) = (10 - 6)/(10 - 4) = 4/6, about 0.667. Equivalently, P(X > 6)/P(X > 4) = 0.4/0.6. The value 0.400 is P(X > 6) without the condition. The value 0.600 is P(X > 4). The value 0.240 multiplies those two probabilities instead of dividing.
Category: Uniform Calculations
Question 10
Which graph shows the probability density function of a uniform distribution on a to b?
A symmetric bell-shaped curve centered at a
A flat horizontal segment from a to b
✓
A curve that starts high and decreases
A hump that is skewed to the right
Correct Answer
A flat horizontal segment from a to b
Every value from a to b is equally likely, so the density is a constant 1/(b - a), drawn as a flat horizontal segment that makes a rectangle with the x-axis. A symmetric bell-shaped curve centered at a describes a normal distribution. A curve that starts high and decreases describes an exponential distribution. A hump that is skewed to the right has unequal heights, so values would not be equally likely.
Category: Uniform Distribution
Question 11
The time between customer arrivals is exponential with a mean of 4 minutes. What is the decay parameter m?
4
16
0.25
✓
0.4
Correct Answer
0.25
For an exponential distribution, m = 1/mu, so m = 1/4 = 0.25 per minute. The value 4 is the mean itself, not its reciprocal. The value 16 is the mean squared, which is the variance. The value 0.4 is a misreading of 1/4 as a decimal.
Category: Exponential Distribution
Question 12
X ~ Exp(1). Use e^(-1) = 0.3679 and e^(-2) = 0.1353 to find P(1 < X < 2), to four decimal places.
0.5032
0.8647
0.6321
0.2326
✓
Correct Answer
0.2326
Since P(X > x) = e^(-mx), the probability between 1 and 2 is e^(-1) - e^(-2) = 0.3679 - 0.1353 = 0.2326. The value 0.5032 adds the two tail areas instead of subtracting. The value 0.8647 is 1 - e^(-2), which is P(X < 2) and includes values below 1. The value 0.6321 is 1 - e^(-1), which is P(X < 1).
Category: Exponential Calculations
Question 13
A waiting time has decay parameter m = 0.5. Given e^(-1) = 0.3679, what is the probability of waiting less than 2 units? Give four decimal places.
0.6321
✓
0.3679
0.5000
0.1840
Correct Answer
0.6321
For an exponential distribution, P(X < x) = 1 - e^(-mx) = 1 - e^(-0.5 times 2) = 1 - 0.3679 = 0.6321. The value 0.3679 is P(X > 2), the area to the right. The value 0.5000 is the decay parameter, not a probability. The value 0.1840 is about the height of the density at 2, which is not an area.
Category: Exponential Calculations
Question 14
Waiting time X is exponential with a mean of 10 minutes. Using ln(2) = 0.693, what is the median of X? Round to two decimal places.
10.00
6.93
✓
5.00
14.43
Correct Answer
6.93
With m = 1/10 = 0.1, the median is ln(2)/m = 0.693/0.1 = 6.93 minutes. The value 10.00 is the mean, which sits above the median because the distribution is skewed right. The value 5.00 assumes the median is simply half the mean. The value 14.43 divides 10 by 0.693 instead of multiplying.
Category: Exponential Calculations
Question 15
Light bulb lifetimes X are exponential. A bulb has already lasted 500 hours. Which expression equals the probability that it lasts at least 100 more hours?
P(X > 600)
P(X > 500)
P(X > 100)
✓
P(X < 100)
Correct Answer
P(X > 100)
The exponential distribution is memoryless: P(X > 500 + 100 | X > 500) = P(X > 100), so a used bulb behaves like a new one. P(X > 600) is the chance a new bulb lasts 600 hours, which ignores that 500 hours have already passed. P(X > 500) is the chance of reaching the age it already has. P(X < 100) is the chance of failing within 100 hours, the complement of the event asked about.
Category: Memoryless Property
Question 16
Times between arrivals at a counter are independent and exponential with a mean of 2 minutes. What is the mean number of arrivals per minute?
2
4
1
0.5
✓
Correct Answer
0.5
When times between events are independent exponential with mean mu, the count of events per unit time is Poisson with mean 1/mu, here 1/2 = 0.5 arrivals per minute. The value 2 is the mean time between arrivals, not the arrival rate. The value 4 squares the mean time. The value 1 would require a mean gap of 1 minute.
Category: Poisson Connection
Question 17
Which random variable is best modeled by an exponential distribution?
Minutes until the next customer arrives
✓
Number of heads in 10 coin flips
Number of defective items in a box
Last digit of a random phone number
Correct Answer
Minutes until the next customer arrives
The exponential distribution models the amount of time until an event occurs, such as waiting for the next customer. The number of heads in 10 coin flips is a discrete count that is binomial. The number of defective items in a box is also a discrete count. The last digit of a random phone number takes ten equally likely whole-number values, a discrete uniform variable.
Category: Exponential Distribution
Question 18
A bus arrives exactly every 15 minutes, and a rider shows up at a random time. Which distribution describes the rider's waiting time?
Exponential with a mean of 15 minutes
Uniform from 0 to 15 minutes
✓
Normal with a mean of 7.5 minutes
Binomial with n = 15 trials
Correct Answer
Uniform from 0 to 15 minutes
Because the rider is equally likely to arrive at any moment within the 15-minute cycle, every waiting time from 0 to 15 minutes is equally likely, which is a uniform distribution. Exponential with a mean of 15 minutes would allow waits longer than 15 minutes, which cannot happen here. Normal with a mean of 7.5 minutes would make waits near 7.5 more likely than others. Binomial with n = 15 trials counts successes and is discrete.
Category: Uniform Distribution
Question 19
For a continuous random variable, what does the cumulative distribution function P(X <= x) represent on the density graph?
The area under the curve to the right of x
The height of the curve directly above x
The area under the curve to the left of x
✓
The area between the mean and x
Correct Answer
The area under the curve to the left of x
The cumulative distribution function accumulates probability from the left, so P(X <= x) is the area under the density to the left of x. The area under the curve to the right of x is P(X > x), which is 1 minus the cumulative value. The height of the curve directly above x is the density f(x). The area between the mean and x leaves out everything below the mean.
Category: Probability as Area
Question 20
For a continuous random variable, P(X < 3) = 0.35. What is P(X >= 3)?
0.35
0.50
0.70
0.65
✓
Correct Answer
0.65
P(X >= 3) is the area to the right of 3, which is 1 - 0.35 = 0.65; including or excluding the single point 3 changes nothing because its probability is 0. The value 0.35 is the area to the left. The value 0.50 assumes the two sides of 3 hold equal area. The value 0.70 doubles the left area, which has no basis.
Category: Probability as Area