Question

5. A college lecturer never finishes his lecture before the end of the hour and always finishes his lecture within 2 min after the hour. Let X= the time that elapses between the end of the hour and the end of the lecture and suppose the pdf of X is
f(x)={kx 2 0 0≤x≤2otherwise f(x)={kx20≤x≤20otherwise
what is the probability that the lecture continues for at least 90 sec beyond the end of the hour?
6. Let X be a continuous rv with cdf
F(x)=⎧ ⎩ ⎨ ⎪ ⎪ 0x4 [1+ln(4x )]0 x≤05≤y≤10x>4 F(x)={0x≤0x4[1+ln(4x)]5≤y≤100x>4. What is the pdf of X?

Answer 1

Answer on Question #59022 – Math – Statistics and Probability

Question

5. A college lecturer never finishes his lecture before the end of the hour and always finishes his lecture within 2 min after the hour. Let $X =$ the time that elapses between the end of the hour and the end of the lecture and suppose the pdf of $X$ is

$$f(x) = \begin{cases} k x^{2}, & 0 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases}$$

what is the probability that the lecture continues for at least 90 sec beyond the end of the hour?

Solution

$$1 = \int_{0}^{2} k x^{2} d x = k \left(\frac{x^{3}}{3}\right)_{0}^{2} = \frac{8}{3} k \rightarrow k = \frac{3}{8}$$ $$P(X \geq 1.5) = \frac{3}{8} \int_{1.5}^{2} x^{2} d x = \frac{3}{8} \left(\frac{x^{3}}{3}\right)_{1.5}^{2} = \frac{2^{3} - 1.5^{3}}{8} = 0.578125 = \frac{37}{64}.$$

Answer: $\frac{37}{64}$.

Question

6. Let $X$ be a continuous rv with cdf

$$F(x) = \begin{cases} 0, & x \leq 0 \\ \frac{x}{4} \left[ 1 + \ln \left(\frac{4}{x}\right) \right], & 0 < x \leq 4 \\ 1, & 1, x > 4 \end{cases}$$

What is the pdf of $X$?

Solution

f(x) = \frac{dF}{dx} = \left\{ \frac{1}{4} \left[ 1 + \ln \left(\frac{4}{x}\right) \right] + \frac{x}{4} \left[ \frac{x}{4} \left(-\frac{4}{x^{2}}\right) \right] = \frac{1}{4} \ln \left(\frac{4}{x}\right), & 0 < x \leq 4, \right.

Answer: $f(x) = \frac{1}{4} \ln \left(\frac{4}{x}\right), 0 < x \leq 4, f(x) = 0$, otherwise.

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