Continuous Random Variables Practice Questions
ATAR (Australia) · ATAR Mathematics Methods · 149 free MCQs with instant results and detailed explanations.
149
Total
52
Easy
68
Medium
29
Hard
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Sample Questions from Continuous Random Variables
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Q1
Easy
What is the expected value of a continuous random variable with a probability density function given by f(x) = 3x^2 for 0 ≤ x ≤ 1?
A.
0.75
B.
1.0
C.
0.5
D.
0.25
Show Answer & Explanation
Correct Answer: A
The expected value E(X) is calculated as the integral of x multiplied by the probability density function over its range. Here, E(X) = ∫ (from 0 to 1) x * (3x^2) dx = 0.75.
Q2
Easy
If the mean of a continuous random variable is 5 and its variance is 4, what is its standard deviation?
A.
2
B.
4
C.
1
D.
5
Show Answer & Explanation
Correct Answer: A
The standard deviation is the square root of the variance. Therefore, if the variance is 4, the standard deviation is √4 = 2.
Q3
Easy
A continuous random variable X has a uniform distribution between 1 and 3. What is the probability that X is less than 2?
A.
0.5
B.
0.25
C.
0.75
D.
0.0
Show Answer & Explanation
Correct Answer: A
In a uniform distribution, the probability can be calculated as the length of the interval where X is less than 2 divided by the total length of the distribution interval. Here, P(X < 2) = (2-1)/(3-1) = 0.5.
Q4
Medium
A continuous random variable X has a probability density function given by f(x) = 3x^2 for 0 ≤ x ≤ 1. What is the probability that X is less than 0.5?
A.
0.125
B.
0.25
C.
0.5
D.
0.75
Show Answer & Explanation
Correct Answer: A
To find P(X < 0.5), we integrate f(x) from 0 to 0.5: P(X < 0.5) = ∫(from 0 to 0.5) 3x^2 dx = [x^3] from 0 to 0.5 = (0.5)^3 - (0)^3 = 0.125.
Q5
Medium
If two continuous random variables X and Y are independent and have probability density functions f_X(x) and f_Y(y), what is the joint probability density function f_{X,Y}(x,y)?
A.
f_X(x) + f_Y(y)
B.
f_X(x) * f_Y(y)
C.
f_X(x) - f_Y(y)
D.
f_X(x) / f_Y(y)
Show Answer & Explanation
Correct Answer: B
For independent continuous random variables, the joint probability density function is the product of their individual densities: f_{X,Y}(x,y) = f_X(x) * f_Y(y).
Q6
Medium
If a continuous random variable X follows a uniform distribution over the interval [2, 8], what is the probability that X is less than 5?
A.
0.375
B.
0.5
C.
0.625
D.
0.75
Show Answer & Explanation
Correct Answer: A
The length of the interval [2, 8] is 6 (8-2). The length of the interval [2, 5] is 3 (5-2). Thus, the probability is 3/6 = 0.5. However, since we need to find the area of the uniform distribution, we calculate the probability as (5-2)/(8-2) = 3/6 = 0.375.
Q7
Medium
A continuous random variable Y has a probability density function given by f(y) = 3y^2 for 0 ≤ y ≤ 1. What is the mean of Y?
A.
0.5
B.
0.25
C.
0.333
D.
0.75
Show Answer & Explanation
Correct Answer: A
The mean of a continuous random variable is computed using the formula E(Y) = ∫(y * f(y)) dy from 0 to 1. This results in E(Y) = ∫(y * 3y^2) dy from 0 to 1, which evaluates to 0.5.
Q8
Hard
A continuous random variable X follows a uniform distribution in the interval [0, 10]. What is the probability that X is less than 4?
A.
0.4
B.
0.6
C.
0.2
D.
0.5
Show Answer & Explanation
Correct Answer: A
The probability of a continuous uniform distribution is calculated as the length of the interval of interest divided by the total length of the distribution interval. Here, the length of the interval [0, 4] is 4, and the total length of the interval [0, 10] is 10. Thus, P(X < 4) = 4/10 = 0.4.
Q9
Hard
The random variable Y is normally distributed with a mean of 50 and a standard deviation of 10. What is the z-score for a value of Y = 65?
A.
1.5
B.
2.0
C.
1.0
D.
0.5
Show Answer & Explanation
Correct Answer: A
The z-score is calculated using the formula z = (Y - mean) / standard deviation. In this case, z = (65 - 50) / 10 = 15 / 10 = 1.5. Therefore, the correct option is A.
Q10
Hard
A continuous random variable X follows a uniform distribution over the interval [a, b]. If the mean of X is 10 and the variance is 25, what are the values of a and b?
A.
0 and 20
B.
5 and 15
C.
10 and 15
D.
5 and 25
Show Answer & Explanation
Correct Answer: B
For a uniform distribution, the mean is (a + b)/2 and the variance is (b - a)²/12. Given the mean is 10, we have a + b = 20. Given variance is 25, we have (b - a)² = 300. Solving these equations leads to a = 5 and b = 15.
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Continuous Random Variables — ATAR (Australia) ATAR Mathematics Methods Practice Questions Online
This page contains 149 practice MCQs for the chapter Continuous Random Variables in ATAR (Australia) ATAR Mathematics Methods. The questions are organized by difficulty — 52 easy, 68 medium, 29 hard — so you can choose the right level for your preparation.
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