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1

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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If $X\sim N(220000, 160000^2)$ describes the profits of the project, then the probability of a loss greater than $80000$ is $\mathbb{P}(X\leq -80000)$. I.e. the probability of a profit less than $-80000$. Now this can be calculated by transforming $X$ into a standard normal random variable $Z$, and looking this probability up in a table as @Ian suggested. Or ...


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$$-80,000 - 220,000=-300,000$$ I.e. an $\$80,000$ loss is $\$300,000$ below the average. $$\dfrac{-\$300,000}{\$160,000} = -1.875 $$ I.e. this is $1.875$ standard deviations below the mean. The probability that a standardized normally distributed random variable is less than $-1.875$ is $\Phi(-1.875)\approx 0.030396$ if I can believe the software I'm ...


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The usual technique is to convert a normal variable $X$, which has mean $m$ and standard deviation $\sigma$, to a normal variable with mean $0$ and standard deviation $1$, usually denoted by $Z$. You do this by defining $Z=\frac{X-m}{\sigma}$. Here $m=220000$ and $\sigma=160000$. Then your inequalities change. Let's say your inequality was $X \leq x$, then ...


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You're thinking in the right direction. Work from first principles (i.e., the very definition of variance).


1

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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He's not comparing the SD of the population to the SD of the sample means. He's saying that the SD of the population of all apple weights is approximately the SD of the sample of 36 apple weights, i.e. $\sigma\approx s$. The intuition is that the histogram of 36 weights should have a similar spread compared to the histogram of all 200,000 weights.


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

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. ...


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The usual model is Poisson, parameter $\lambda=5$. The probability of more than $3$ blackouts is $1$ minus the probability of $\le 3$ blackouts. The probability of $\le 3$ blackouts is $$e^{-5}\left(1+\frac{5}{1!}+\frac{5^2}{2!}++\frac{5^3}{3!}\right).$$ The mean monthly cost is indeed $7500$. If $X$ is the number of blackouts, then the cost $Y$ is ...


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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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Your second equation for $\text{var}(XY)$ is true if $X$ and $Y$ are independent, not in general otherwise. I don't know what kind of an answer you're expecting for $\text{var}(X/Y)$. There is no formula expressing $E[1/Y]$ or $\text{var}(1/Y)$ in terms of $E[Y]$ and $\text{var}(Y)$, if that's what you want. EDIT: One thing you can say is this. Suppose ...



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