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Gary: Mean: The sum of elements divided by the number of elements in a set. (0 + 20)/2 = 10 Median: The data set only contains two elements so the median is found by taking the mean of those elements, e.g. it’s 10. Range: 20-0 = 0.


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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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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 $k=2$.


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The second one is the right one since $\int_1^2 x-1 \, dx=0.5$. If we add $F\left(1^+ \right)=0.5$ we get $1$. It fulfills one property of a valid pdf: $\int\limits_{-\infty}^{\infty} f(x) \, dx=1$


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


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Generally speaking, NNs are better at classification than regression, so I'm assuming the ground truth is something like $[1,0]$ for the cars and $[0,1]$ for everything else (easily done with something like keras.utils.to_categorical) and your wanted output is $[P,Q]$ where $P$ is the confidence of "this is a car", $Q$ is the confidence of "this is not a car"...


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There might be better ways from people who have actually studied such questions, but I'd do this as follows. I'm assuming, as in your example, that you get answers in the the range from 0 to 100, 100 meaning definite "yes" to the question, 50 meaning "don't know" and 0 means a definite "no" to the question. Then do the following 1) Order the results ($p_1,...


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Infinite mean and mean not existing are not one and the same. $X$ has mean $\infty$ if $EX^{+}=\infty$ and $EX^{-}<\infty$. Mean of $X$ dose not exist if $EX^{+}=\infty$ and $EX^{-}=\infty$ in which case you cannot assign any value, finite of infinite, to $EX$. We can say the following: if $EX$ exists (finite or infinite) and $E(X-EX)^{2} <\infty$ ...


2

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 ...


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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 ...


0

You can calculate the individual probabilities: $$P(\text{n spades in 5 draws})=\frac{{39 \choose 5-n}{13 \choose n}}{{52 \choose 5}}$$ So the expected value would be $$E(\text{# of spades})=\sum_{n=0}^{5}n.P(n)=\sum_{n=0}^{5}n.\frac{{39 \choose 5-n}{13 \choose n}}{{52 \choose 5}}$$


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Hint: first, try this problem with selecting 2 cards out of 52. What is the probability of selecting 0 spades out of those 2? $P(0) = (39/52)*(38/51) = 4/17$. What is the probability of selecting 1 spade out of those 2, $P(1)$? It's the probability of selecting a spade followed by a non-spade, plus the probability of selecting a non-spade followed by a ...


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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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I will offer some clues and comments because this problem is either unclear or extremely unrealistic, and alleged answers are puzzling. It seems you already have $n = 30,$ sample mean $$\bar X = \frac 1 n\sum_i X_i = 243/30 = 8.1$$ and sample standard deviation $$S = \sqrt{\frac{1}{n}\left[\sum_i X^2 - (\sum_i X)^2/n\right]} = 3.41,$$ although the usual ...


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