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For $n\geq k>0$ we obtain: $$a^n+b^n-a^kb^{n-k}-a^{n-k}b^k=(a^k-b^k)(a^{n-k}-b^{n-k})\geq0.$$ For $0<n<k$ it's wrong. Try, $n=1$ and $k=2$.

2

$a^5+b^5+c^5+a^5+b^5>5abcab$ $a^5+b^5+c^5+b^5+c^5>5abcbc$ $a^5+b^5+c^5+a^5+c^5>5abcac$ Strict inequalities hold since $a$, $b$ and $c$ are distinct. Add them up and the result follows.

2

Mean weight and median weight are just the mean and median. In this case we're talking about weight. We could say mean age, mean height, mean length, if we were talking about something else. The mean is the numerical average, while the median is the middle element in an ordered data set.

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Use the additivity of expectation: expectation of sum of RVs is the sum of their expectations. Let $X_i,~i=1,\dots,5$ be RVs with $X_i = 1$ if the $i$-th card is a spade and $=0$ otherwise. Then these RVs are identically distributed (but not independent), and each is Bernoulli with parameter $1/4$ (fourth of the cards are spades), so $E[X_i] = 1/4$. Why ...

2

You are correct (your notation could be more precise and clearer). By the Cauchy-Schwarz inequality, we have that $$\operatorname E|X_n-X|\le\sqrt{\operatorname E|X_n-X|^2}\to0$$ as $n\to\infty$, which shows that convergence in mean square implies convergence in mean, i.e. $\operatorname E|X_n-X|\to0$ as $n\to\infty$.

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1) I'm not sure about standard references. But the usual definition Sort all items by their value x. The median is the value of the $\frac{N+1}{2}$th item. is excessively procedural, and refers to a slower-than-needed algorithm. A better definition nearby is: The median is an $x$ such that at least half the values are $\ge x$, and at least half the ...

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Hint: For $k = 1,2,3,4,5$, let $X_k$ be $1$ if the $k$-th card you drew is a spade, and $0$ otherwise. Then, $$X = X_1+X_2+X_3+X_4+X_5$$ and then by using linearity of expectation, $$E[X] = E[X_1+X_2+X_3+X_4+X_5] = E[X_1]+E[X_2]+E[X_3]+E[X_4]+E[X_5].$$ Also, for each $k$, we have E[X_k] = 1 \cdot P[X_k = 1] + 0 \cdot P[X_k = 0] = P[\text{the} \ k\text{-th ...

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