Mathematics Week 12 Graded Assignment
Graded Assignment
1. Uniform Distribution Expectation and Variance
Answer: $ab = e^2 - 3v$ Solution:
- Expectation: $E[X] = \frac{a + b}{2} = e \implies (a + b)^2 = 4e^2$
- Variance: $\text{Var}(X) = \frac{(b - a)^2}{12} = v \implies (b - a)^2 = 12v$
- Subtract: $(a + b)^2 - (b - a)^2 = 4ab = 4e^2 - 12v \implies ab = e^2 - 3v$1.
Answer: $n^3 - m^3$ Solution:
- Normalize: $\int_0^1 kx^2 dx = 1 \implies k = 3$.
- Probability: $P(m < X < n) = \int_m^n 3x^2 dx = n^3 - m^3$1.
Answer: $\frac{120 - m}{120 - n}$ Solution:
- Conditional probability: $P(X > m \mid X > n) = \frac{1 - \frac{m - 100}{20}}{1 - \frac{n - 100}{20}} = \frac{120 - m}{120 - n}$1.
Answer: $1 - (1 - e^{-8\lambda})^5 - 5e^{-8\lambda}(1 - e^{-8\lambda})^4$ Solution:
- Probability a customer waits >8 minutes: $e^{-8\lambda}$.
- Use binomial distribution: $P(\geq 2) = 1 - P(0) - P(1)$1.
Answer: 0.425 Solution:
- Integrate piecewise: $\int_0^1 0.2x , dx + \int_1^2 0.2 , dx + \int_2^{2.5} (0.2x - 0.2) , dx = 0.425$1.
Answer: $2t$ Solution:
- Variance = $t^2 \implies \lambda = 1/t$.
- Expected time for two arrivals: $2 \cdot \frac{1}{\lambda} = 2t$1.
Answer: $e = \frac{a + b}{2} = \frac{12 + 2a}{2}$ Solution:
- From $(b - a)^2 = 144$ and $m = a + \frac{12}{p}$1.
Answer: $t = n \ln\left(\frac{100}{100 - p}\right)$ Solution:
- Solve $1 - e^{-t/n} = \frac{p}{100} \implies t = n \ln\left(\frac{100}{100 - p}\right)$1.
Answer: Yes Solution:
- Use two-sample t-test; reject $H_0$ if $|t| > 1.96$2.
Answer: $\frac{p}{\alpha}$ Example: For $p = 1/3, \alpha = 3/4$, bound = $4/9$3.
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https://www.studocu.com/in/document/indian-institute-of-technology-madras/iitm-online-degree-data-science-and-programming/week-12-graded-solution/102388299 ↩︎ ↩︎ ↩︎ ↩︎ ↩︎ ↩︎ ↩︎ ↩︎
https://www.scribd.com/document/640063296/Stats-2-Week-12-GA-Studify-space ↩︎
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