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Tutor profile: Adrian K.

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Adrian K.
Tutor for 4 Years, PhD Candidate in Statistics
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Questions

Subject: Pre-Calculus

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Question:

Consider the quadratic equation $f(x) = ax^2 + bx - c$ where $a, b, c$ are all positive constants. Show there exists at least one real root.

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Adrian K.
Answer:

To show there exists at least one real root, we simply need to show the graph is zero for some value of $x$. First note that by construction, $f(0) = a\times 0^2 + b\times 0 - c= - c < 0$. So that the function is strictly negative at $x = 0$. Similarly, note that since $a$ and $b$ are positive, the function goes to infinity as $x$ goes to infinity, $\lim_{x\to \infty} f(x) = \lim_{x\to \infty} ax^2 + bx - c = \infty$ So we have shown that $f(x) = 0$ at $x=0$ and goes to positive infinity as $x \to \infty$. Since $f(x)$ is clearly continuous, this implies it must cross the $x$ axis at some point, and hence a real root exists.

Subject: Pre-Algebra

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Question:

An apple costs $x$ dollars. A watermelon is twice as expensive as an apple. A banana is $\frac{2}{3}$ as expensive as a watermelon. How expensive is a banana compared to an apple?

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Adrian K.
Answer:

Let $y$ be the cost of a pineapple, and $z$ the cost of a banana. We wish to find the number $A$ such that $z = A x$. We are told a watermelon is twice as expensive as an apple. This can be written mathematically as $y = 2x$. Similarly, we are told a banana is two-thirds as expensive as a pineapple, and this can be written mathematically as $z = \frac{2}{3} y$ To find $A$, we simply substitute the first equation for $y$ (in terms of $x$) in the second equation. This gives $z = \frac{2}{3} y = \frac{2}{3} \times 2x = \frac{4}{3} x$. Thus, a banana is four-thirds as expensive as an apple.

Subject: Statistics

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Question:

Let $X_1, X_2, \dots, X_n$ be a random sample from an unknown distribution with finite mean $\mu$ and variance $\sigma^2$. Find the asymptotic distribution of $\sqrt{n}s_n^2$ where $s_n^2=\frac{1}{n}\sum_{i=1}^n (X_i - \bar{X}_n)^2$.

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Adrian K.
Answer:

We write $\sum_{i=1}^n( X_i-\bar{X}_n)^2 = \sum_{i=1}^n (X_i-\mu)^2 + n(\bar{X}_n-\mu)^2$ The left hand term is the summation of identically distributed random variables. Thus we have $\sqrt{n}s_n^2 = \frac{1}{\sqrt{n}}\sum_{i=1}^n (X_i-\mu)^2 + \sqrt{n}(\bar{X}_n-\mu)^2$ The left-hand term is asymptotically normal by the central limit theorem. The right hand term converges to zero in probability since we can write $\sqrt{n}(\bar{X}_n-\mu)^2 = \frac{1}{\sqrt{n}} (\sqrt{n}(\bar{X}_n-\mu))^2$ and $\sqrt{n}(\bar{X}_n-\mu)$ is asymptotically normal by the central limit theorem. It then follows by Slutsky's theorem that $\sqrt{n}s_n^2$ is asymptotically normal.

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