Q1: A continuous random variable differs from a discrete random variable in that it: (a) Can take any value within a given interval (b) Can only take whole number values (c) Has a finite number of possible outcomes (d) Cannot be measured precisely
Solution:
Ans: (a) Explanation: A continuous random variable can assume any value within a specified interval or range, unlike discrete random variables which take on countable values. Examples include height, weight, and time.
Q2: For a continuous random variable, the probability of it taking on any single exact value is: (a) 0.5 (b) 1 (c) 0 (d) Depends on the value
Solution:
Ans: (c) Explanation: For continuous random variables, \(P(X = a) = 0\) for any specific value \(a\). Instead, we calculate probabilities over intervals using \(P(a < x="">< b)\).="" this="" is="" because="" there="" are="" infinitely="" many="" possible="" values="" in="" any="" interval.="">
Q3: The total area under a probability density function (PDF) curve must equal: (a) 0 (b) 0.5 (c) 1 (d) Infinity
Solution:
Ans: (c) Explanation: The probability density function must satisfy the property that the total area under the curve equals 1, representing the certainty that the random variable will take some value within its range. This is expressed as \(\int_{-\infty}^{\infty} f(x) \, dx = 1\).
Q4: If X is a continuous random variable with PDF f(x), then P(a ≤ X ≤ b) is found by: (a) Adding f(a) and f(b) (b) Calculating the area under f(x) from a to b (c) Multiplying f(a) by f(b) (d) Subtracting f(a) from f(b)
Solution:
Ans: (b) Explanation: The probability \(P(a \leq X \leq b)\) is calculated as the area under the PDF curve between \(a\) and \(b\), which is found using integration: \(\int_{a}^{b} f(x) \, dx\). The PDF values themselves are not probabilities.
Q5: For a uniform distribution on the interval [2, 8], the probability density function f(x) equals: (a) 1/6 (b) 1/8 (c) 1/10 (d) 1/2
Solution:
Ans: (a) Explanation: For a uniform distribution on interval \([a, b]\), the PDF is \(f(x) = \frac{1}{b-a}\). Here, \(f(x) = \frac{1}{8-2} = \frac{1}{6}\) for \(2 \leq x \leq 8\), and 0 elsewhere.
Q6: The cumulative distribution function (CDF) F(x) of a continuous random variable represents: (a) The probability that X equals x (b) The probability that X is greater than x (c) The probability that X is less than or equal to x (d) The derivative of the PDF at x
Solution:
Ans: (c) Explanation: The cumulative distribution function \(F(x) = P(X \leq x)\) gives the probability that the random variable takes a value less than or equal to \(x\). It is calculated as \(F(x) = \int_{-\infty}^{x} f(t) \, dt\).
Q7: If the CDF of a continuous random variable is F(x), then the PDF f(x) can be found by: (a) Integrating F(x) (b) Differentiating F(x) (c) Taking the square root of F(x) (d) Adding 1 to F(x)
Solution:
Ans: (b) Explanation: The probability density function is the derivative of the CDF: \(f(x) = \frac{d}{dx}F(x)\). This relationship shows that the PDF represents the rate of change of cumulative probability.
Q8: For a continuous random variable X with PDF f(x), which statement is always true? (a) f(x) ≥ 0 for all x (b) f(x) ≤ 1 for all x (c) f(x) = P(X = x) (d) The maximum value of f(x) is 1
Solution:
Ans: (a) Explanation: A valid PDF must satisfy \(f(x) \geq 0\) for all values of \(x\). While the total area under \(f(x)\) must equal 1, the function values themselves can exceed 1. The PDF does not give probabilities directly, only probability densities.
## Section B: Fill in the Blanks
Q9: A random variable that can take any value within an interval is called a __________ random variable.
Solution:
Ans: continuous Explanation:Continuous random variables can assume infinitely many values within a given range, as opposed to discrete random variables which take specific, countable values.
Q10: The function that describes the probability distribution of a continuous random variable is called the __________ __________ function.
Solution:
Ans: probability density Explanation: The probability density function (PDF) characterizes continuous random variables. The probability of an interval is found by integrating the PDF over that interval.
Q11: For any continuous random variable X, P(X = c) = __________ for any constant c.
Solution:
Ans: 0 Explanation: Since continuous random variables have infinitely many possible values, the probability of any single exact value is zero. Only intervals have non-zero probabilities.
Q12: The cumulative distribution function F(x) is defined as F(x) = P(X __________ x).
Solution:
Ans: ≤ Explanation: The CDF gives the probability that the random variable is less than or equal to a specific value: \(F(x) = P(X \leq x)\).
Q13: If f(x) is a valid probability density function, then \(\int_{-\infty}^{\infty} f(x) \, dx\) = __________.
Solution:
Ans: 1 Explanation: A fundamental property of any PDF is that the total area under the curve equals 1, representing the total probability of all possible outcomes.
Q14: The relationship between the PDF f(x) and CDF F(x) is given by f(x) = __________ F(x).
Solution:
Ans: \(\frac{d}{dx}\) or F'(x) or derivative of Explanation: The PDF is the derivative of the CDF with respect to \(x\): \(f(x) = \frac{d}{dx}F(x)\). Conversely, the CDF is the integral of the PDF.
## Section C: Word Problems
Q15: A bus arrives at a stop uniformly between 2:00 PM and 2:20 PM. If you arrive at 2:00 PM, what is the probability that you will wait more than 15 minutes for the bus?
Solution:
Ans:
The bus arrival time follows a uniform distribution on [0, 20] minutes.
For a uniform distribution, \(P(X > a) = \frac{b - a}{b - c}\) where the interval is [c, b].
We want \(P(X > 15)\) where \(X\) is uniformly distributed on [0, 20].
\(P(X > 15) = \frac{20 - 15}{20 - 0} = \frac{5}{20} = \frac{1}{4} = 0.25\) Final Answer: 0.25 or 1/4 or 25%
Q16: The time (in hours) it takes for a battery to charge is a continuous random variable with PDF \(f(x) = \frac{1}{4}\) for \(0 \leq x \leq 4\) and 0 otherwise. Find the probability that the battery takes between 1 and 3 hours to charge.
Solution:
Ans:
We need to find \(P(1 \leq X \leq 3)\).
\(P(1 \leq X \leq 3) = \int_{1}^{3} f(x) \, dx = \int_{1}^{3} \frac{1}{4} \, dx\)
\(= \frac{1}{4}[x]_{1}^{3} = \frac{1}{4}(3 - 1) = \frac{1}{4}(2) = \frac{1}{2}\) Final Answer: 0.5 or 1/2
Q17: A continuous random variable X has the PDF \(f(x) = kx\) for \(0 \leq x \leq 2\) and 0 elsewhere. Find the value of the constant k that makes this a valid probability density function.
Solution:
Ans:
For a valid PDF, \(\int_{-\infty}^{\infty} f(x) \, dx = 1\).
\(\int_{0}^{2} kx \, dx = 1\)
\(k \int_{0}^{2} x \, dx = 1\)
\(k \left[\frac{x^2}{2}\right]_{0}^{2} = 1\)
\(k \left(\frac{4}{2} - 0\right) = 1\)
\(2k = 1\)
\(k = \frac{1}{2}\) Final Answer: k = 1/2 or 0.5
Q18: The amount of coffee (in ounces) dispensed by a machine is uniformly distributed between 7.5 and 8.5 ounces. What is the probability that a randomly selected cup contains between 7.8 and 8.2 ounces?
Solution:
Ans:
The coffee amount follows a uniform distribution on [7.5, 8.5].
For a uniform distribution on [a, b], \(P(c \leq X \leq d) = \frac{d - c}{b - a}\).
\(P(7.8 \leq X \leq 8.2) = \frac{8.2 - 7.8}{8.5 - 7.5} = \frac{0.4}{1.0} = 0.4\) Final Answer: 0.4 or 2/5
Q19: A continuous random variable has CDF \(F(x) = \frac{x^2}{16}\) for \(0 \leq x \leq 4\). Find the probability density function f(x).
Solution:
Ans:
The PDF is found by differentiating the CDF: \(f(x) = \frac{d}{dx}F(x)\).
\(f(x) = \frac{d}{dx}\left(\frac{x^2}{16}\right)\)
\(f(x) = \frac{2x}{16} = \frac{x}{8}\)
Therefore, \(f(x) = \frac{x}{8}\) for \(0 \leq x \leq 4\) and 0 elsewhere. Final Answer: \(f(x) = \frac{x}{8}\) for \(0 \leq x \leq 4\)
Q20: The waiting time (in minutes) at a doctor's office is uniformly distributed between 5 and 25 minutes. Find the probability that a patient waits at most 12 minutes.
Solution:
Ans:
The waiting time follows a uniform distribution on [5, 25].
We need \(P(X \leq 12)\).
For a uniform distribution, \(P(X \leq c) = \frac{c - a}{b - a}\) where the interval is [a, b].
\(P(X \leq 12) = \frac{12 - 5}{25 - 5} = \frac{7}{20} = 0.35\) Final Answer: 0.35 or 7/20
The document Worksheet (with Solutions): Continuous Random Variables is a part of the Grade 9 Course Statistics & Probability.
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