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If you have a discrete random variable $X$ that takes on values $x_i$ with probabilities $p(x_i), 1 \leq i \leq n$, and a function $f(x)$, then the expected value of $f(x)$ is given by $$E(f(X)) = \sum_{i=1}^n f(x_i) p(x_i)$$ That is LOTUS. Perhaps an example will make it clearer. Suppose that $X$ takes on values $1, 2, 3$ with probabilities $1/2, 1/3, ... 2 For a discrete random variable$X$taking on values$x_1, x_2, \ldots,x_n$with probabilities$p_1, p_2, \ldots, p_n$respectively, the expected value or expectation of$X$, commonly denoted as$E[X]$, is defined to be $$E[X] = \sum_{i=1}^n x_i p_i.\tag{1}$$ (Feel free to replace each$p_i$with$p(x_i)$if that feels more comfortable). Similarly, the ... 2 (a) and (b) are OK. (c) 0.9996645, so a slight rounding error at worst. (d) In (c) you are using the PDF$F(x) = 1 - e^{-2x},$for$x > 0$to find the answer. Here you need$F(5) - F(2),$which requires no additional integration. Just watch the signs and you'll get something close to .02. It should be easy to make a graph of the exponential density ... 1 Chebyshev's Inequality is true for any$c > 0$, but you are right that it only provides useful information for$c > 1$. This is actually surprisingly easy prove. Define$\mu = E(X)$and$\sigma^2 = E((X-\mu)^2)$. Observe that for any$c \geq 0$we have$\mathbb{1}\left\{\left|\frac{X-\mu}{\sigma}\right|\geq c\right\} \leq \frac{(X-\mu)^2}{\sigma^2 ...

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Your second line should be $$=\frac{1}{n-1}\left [\sum_{i=1}^nx_i^2+n\bar{x}^2-2\bar{x}\sum_{i=1}^n x_i \right ]$$ Then just as you said, you can use $$\sum_{i=1}^n\frac{x_i}{n}=\bar{x}$$ to continue.

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In your confidence interval, you are using $\bar x = 7$, which is inconsistent with your earlier notation. It should be $\bar d = \bar y - \bar x= 7.$ Also, to be clear, 16 is the variance for the difference in scores for one individual, so the variance for $\bar d$ is $16/3$ and the standard deviation used in the confidence interval is $4/\sqrt{3},$ as you ...

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I'm assuming a very large population of units from which to select, so that the distribution is binomial, and that 'have issues' means 'are defective." Then we have the number of defective units seen $X \sim Bin(n=10, p=.1)$ Thus, (1) $E(X) = np = 1$ as you say. (2) V(X) = np(1-p) = .9, so SD(X) = \sqrt(.9) = 0.9487.$; I think you overlooked a decimal point. ... 1 This looks like you are supposed to assume such accidents are a rare event. The question also states "accidents occur at random times and independently from one another.". These both hint that we should model the data using a Poisson distribution,$\operatorname{Po}(\lambda)$.$\lambda$is the mean, so$\lambda=15\$. Of course, the Poisson distribution also ...

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