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Tutor profile: Armen P.

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Armen P.
Professor at a top 10 US university
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Questions

Subject: Applied Mathematics

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Question:

You flip a coin 5 times, two heads come up. Then you give it to a friend and he flips it 6 times until a head comes up on the 7-th. Let $$\theta$$ be the probability of the coin coming up the head. Compute the Estimate $$\theta$$.

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Armen P.
Answer:

We will use a Bayes estimator. Let $$X_i=1$$ when it is a head on the $$i$$-th trial and 0 if it is a tail. Then $$X$$ has a binomial distribution with parameter $$\theta$$. Our data is given in the form of 12 trials with 3 successes. Then, from the Bayes estimate formula for the mean of Binomial distribution, we have $$\bar \theta=\frac{1}{n}\sum\limits_{i=1}^{12}X_i=\frac{3}{12}=\frac{1}{4}.$$

Subject: Calculus

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Question:

Find the value of c for which the function $$f(x)=c\cdot (1-x^2), x\in [0,1]$$ integrate to 1.

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Armen P.
Answer:

$$1=\int_{-\infty}^\infty f(x)\de x=\int_{0}^1c\cdot (1-x^2)\de x=c\cdot (x-\frac{x^3}{3})\bigg |_{x=0}^1=c\cdot (1-\frac{1}{3})=c\cdot\frac{2}{3}.$$ therefore $$c=3/2$$.

Subject: Statistics

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Question:

A point is uniformly and repeatedly thrown inside the interval $$[0,1]$$. What is the smallest number of throws needed so that the probability of the point landing inside $$[0.4, 0.6]$$ at least once exceeds $$0.95$$.

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Armen P.
Answer:

We think of a success when the point lands in the interval $$[0.4, 0.6]$$. The probability of success is $$p=0.6-0.4=0.2.$$ Let X be the number of throws until the first success happens. Then X is geometrically distributed and we want to compute $$P(X\leq n)=1-(1-p)^n\geq 0.95\Rightarrow (0.8)^n\leq 0.05\Rightarrow n\ln 0.8\leq 0.05\Rightarrow n\geq \frac{\ln 0.05}{\ln 0.8}\approx 13.4251 $$ Where we used the formula of cdf of a geometric distribution. Hence the smallest n is equal to 14.

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