For a group of students, the time taken to get to school, in minutes, is a continuous random variable, [math: T] . The probability distribution for [math: T] is given by: [math: f(t)={−14500t(t−30),0≤t≤300,elsewhere] f(t)= \begin{cases} -\frac{1}{4500} t(t-30)\:\:,\:\: 0 \leq{t} \leq30 \ \:\:\:\:\:0\:\:,\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:elsewhere \end{cases} .
It is known that the slowest 10% of students take more than [math: k] minutes to get to school, i.e. [math: Pr(Tk)=0.1] Which of the following is closest to the value of [math: k] ?

Question

For a group of students, the time taken to get to school, in minutes, is a continuous random variable, [math: T] .  The probability distribution for [math: T] is given by: [math: f(t)={−14500t(t−30),0≤t≤300,elsewhere] f(t)= \begin{cases} -\frac{1}{4500} t(t-30)\:\:,\:\: 0 \leq{t} \leq30 \ \:\:\:\:\:0\:\:,\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:elsewhere \end{cases} .
It is known that the slowest 10% of students take more than [math: k] minutes to get to school, i.e. [math: Pr(T>k)=0.1] Which of the following is closest to the value of [math: k] ?

BlackTom AI · Accepted Answer

We start by restating the setup and the given probability density function (PDF) for T, the time to get to school.
- The PDF is f(t) = (-14500t(t−30)) / 1, for 0 ≤ t ≤ 30, and 0 elsewhere. This is equivalently written as f(t) = ( -t^2 + 30t ) / 4500 on 0 ≤ t ≤ 30. The function is zero outside the interval [0,30].
- First, verify it integrates to 1 to ensure it is a valid PDF: ∫ from 0 to 30 of ( -t^2 + 30t ) / 4500 dt = (1/4500)[ -t^3/3 + 15t^2 ] from 0 to 30. Plugging t = 30 gives (1/4500)[ -27000/3 + 15·900 ] = (1/4500)[ -9000 + 13500 ] = 1. So the PDF is valid.
Next, we derive the CDF F(t) = P(T ≤ t) for 0 ≤ t ≤ 30:
- F(t) = ∫ from 0 to t of ( -u^2 + 30u ) / 4500 du = (1/4500)[ -t^3/3 + 15t^2 ].
- Therefore, P(T > t) = 1 − F(t) = 1 − (1/4500)[ -t^3/3 + 15t^2 ].
We are told that the slowest 10% take more than k minutes, i.e. P(T > k) = 0.1. Equivalently, P(T ≤ k) = 0.9, so F(k) = 0.9. This gives:
(1/4500)[ -k^3/3 + 15k^2 ] = 0.9.
Multiplying both sides by 4500 yields: -k^3/3 + 15k^2 = 4050. Multiplying through by 3 to clear the fraction gives: -k^3 + 45k^2 = 12150, or equivalently k^3 − 45k^2 + 12150 = 0.
Now we solve the cubic k^3 − 45k^2 + 12150 = 0 for k in the interval [0,30].
- Evaluate the left-hand side (call it G(k)) at the candidate values: G(21) = 21^3 − 45·21^2 + 12150 = 9261 − 19845 + 12150 ≈ 566 (positive); G(24) = 13824 − 25920 + 12150 = 54 (close to zero); G(28) = 21952 − 35280 + 12150 ≈ −178? (negative). This shows the root lies between 24 and 28, and closer to 24.
- To refine, consider a couple of near-24 values: for k = 24.0, G ≈ 54; for k = 24.2, G ≈ −28; for k = 24.1, G is very close to zero (slightly positive). This indicates the root is approximately 24.1 minutes, i.e., the value of k that satisfies P(T > k) = 0.1 is around 24 minutes.
Conclusion from the calculation: the value of k that yields P(T > k) = 0.1 is near 24 minutes. Among the provided options, the one closest to this computed value is the choice that corresponds to about 24 minutes. 
Notes on the other options to clarify common pitfalls:
- A. 21 minutes would imply F(21) = 0.9 only if the cubic equation were exactly satisfied at 21; however, plugging k = 21 gives G(21) ≈ 566, which is not near zero, so 21 is not a good approximation.
- B. 6 minutes sits far from the root region (the CDF at 6 is small, so P(T ≤ 6) would be far less than 0.9), making it an inaccurate estimate for the 90th percentile.
- C. 28 minutes lies further to the right in the interval and results in G(28) being noticeably negative, indicating it overshoots the root and underestimates P(T ≤ k) relative to 0.9.
- D. 24 minutes is the value that makes the cubic expression closest to zero, aligning with the required F(k) = 0.9 and P(T > k) = 0.1 within the given PDF support.
In summary, using the given PDF on [0,30], the condition P(T > k) = 0.1 translates to solving a cubic that yields a root near 24 minutes, making the option near 24 minutes the closest match among the choices.

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