Catalogue DP Mathematics Mathematics: Analysis and Approaches HL Statistics and probability Continuous random variables

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Continuous random variables Practice

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[[input-continuous-random-variables-practice-7]]","widgets":{"input-continuous-random-variables-practice-7":{"type":"input","answer":"0"}},"hints":[],"explanation":"The density decreases, so its maximum is at the left endpoint.","id":"continuous-random-variables-practice-7","evidence_kind":"practice","correct_feedback":"The density decreases, so its maximum is at the left endpoint.","incorrect_feedback":"The mode is where the density is largest, not where half the probability lies. This density decreases across its support."},{"content":"A student rejects density 4 on [0,0.25] because it exceeds 1. 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[[input-continuous-random-variables-practice-9]]","widgets":{"input-continuous-random-variables-practice-9":{"type":"input","answer":"0.875"}},"hints":[],"explanation":"The cumulative probability is x³, so subtract 0.5³ from 1.","id":"continuous-random-variables-practice-9","evidence_kind":"practice","correct_feedback":"The cumulative probability is x³, so subtract 0.5³ from 1.","incorrect_feedback":"The interval probability is cumulative area at 1 minus cumulative area at 0.5. Integrate 3x² to obtain x³ before substituting."},{"content":"For $f(x)=2x$ on [0,1], find $P(X\u003E0.75\\mid X\u003E0.5)$.\u003Cbr\u003E[[choice-continuous-random-variables-practice-10]]","widgets":{"choice-continuous-random-variables-practice-10":{"type":"choice","options":["$7/12$","$7/16$","$3/4$"],"answer":[0],"choice_feedback":["The exact ratio is (1 − 0.75²)/(1 − 0.5²) = 7/12.","7/16 is the unconditional area above 0.75. 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Practice explanations

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For a continuous variable X, what is $P(X=1.7)$? Enter an integer.

Correct answer: 0

A single point has no area under a density.

A density is $f(x)=k$ on [2,6] and zero elsewhere. Enter k as a decimal.

Correct answer: 0.25

Total area is 4k = 1.

A density is $f(x)=kx^2$ on [0,3] and zero elsewhere. Find k.

$1/9$: The integral of x² from 0 to 3 is 9, so 9k = 1.
$1/3$: Integrating x² produces x³/3. The integral over [0,3] is 9, so a factor of 1/3 would give too much total area.
$1/27$: This omits the division by 3 in the antiderivative. The area under x² is 9 rather than 27.

The integral of x² from 0 to 3 is 9, so 9k = 1.

For $f(x)=2x$ on [0,1], enter $P(X<0.5)$ as a decimal.

Correct answer: 0.25

Probability is area, not density height. Integrating 2x gives x², so the area up to 0.5 is 0.25.

For f(x) = 2x on [0,1], the shaded area up to 0.5 is 0.25. Move the upper bound to compare area with height.

For $f(x)=2x$ on [0,1], find $E(X)$.

$2/3$: Integrate x(2x) over [0,1], giving 2/3.
$1/2$: The midpoint 1/2 is the mean of a uniform density. Here higher values have more density, so the weighted mean lies to its right.
$1$: The density peaks at 1, but the mode is not the mean. Expectation averages all values using their density weights.

Integrate x(2x) over [0,1], giving 2/3.

For $f(x)=2x$ on [0,1], find the median.

$1/\sqrt{2}$: The cumulative probability is x²; set m² = 1/2 and use the positive root.
$1/2$: Half the support length need not contain half the probability. Here the area up to 1/2 is only 1/4.
$2/3$: 2/3 is the mean, found by a weighted integral. The median instead makes the cumulative area equal 1/2.

The cumulative probability is x²; set m² = 1/2 and use the positive root.

Explain why

The median m satisfies $\int_0^m2x\,dx=m^2=1/2$. Choose the positive root within the support. The split is between equal areas, not equal horizontal lengths.

Initially the shaded area is exactly half of the total, even though the upper bound is greater than 0.5. This is the median, 1/√2.

For $f(x)=2(1-x)$ on [0,1], enter the mode.

Correct answer: 0

The density decreases, so its maximum is at the left endpoint.

A student rejects density 4 on [0,0.25] because it exceeds 1. Is the density valid?

Yes, its total area is 1 and it is nonnegative: Only probabilities, which are areas, must be at most 1.
No, density heights must be at most 1: A density may exceed 1 when its support is narrow. The constraint is on total area, which here is 4 × 0.25 = 1.
No, its mean is greater than 1: Every possible value is between 0 and 0.25, so the mean cannot exceed 1. Check normalization rather than imposing a bound on density height.

Only probabilities, which are areas, must be at most 1.

The density has height 4 over width 0.25, giving total area 1. A height above 1 is valid; probability is area.

For $f(x)=3x^2$ on [0,1], enter $P(0.5<X<1)$ as a decimal.

Correct answer: 0.875

The cumulative probability is x³, so subtract 0.5³ from 1.

For $f(x)=2x$ on [0,1], find $P(X>0.75\mid X>0.5)$.

$7/12$: The exact ratio is (1 − 0.75²)/(1 − 0.5²) = 7/12.
$7/16$: 7/16 is the unconditional area above 0.75. Once X > 0.5 is known, divide by the remaining area 3/4.
$3/4$: 3/4 is the probability of the condition X > 0.5. It is the new denominator, not the conditional probability asked for.

The exact ratio is (1 − 0.75²)/(1 − 0.5²) = 7/12.

Explain why

Conditioning on $X>0.5$ removes the probability below 0.5. The desired area above 0.75 is then measured as a fraction of that remaining probability, not of the original total.

The hatched region is the conditioning area X > 0.5, totaling 0.75. The darker selected area X > 0.75 totals 0.4375. Their ratio is 7/12.

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