# Tag Info

1

As commented on the question, the assumption has to be made that Kyle is the same number of standard deviation units from the mean. Therefore my calculation is correct. If that assumption cannot be made then the probably is not possible to solve with the given information.

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When you calculate the p do not round up, instead it'll just be (55%*24)/100=13.33 which is the correct answer.

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My answer based on the second version of the original post: According to the central limit theorem, the standard deviation of the sample mean of $n$ data from a population is $\sigma_{\overline{X}}=\sigma_X/\sqrt{n}$, where $\sigma_X$ is the population standard deviation. In your case, $\sigma_{\overline{X}}=40/\sqrt{100}=4$. My answer based on the first ...

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Generally speaking, in the Black-Scholes model, the stock price process $\{S(t), t \geq 0\}$ is modelled as $$S(t) = S(0) e^{\displaystyle \left(r-\frac{1}{2}\sigma^2\right)t + \sigma W(t)}.$$ Now, comparing the above with $S(t) = 6e^{-2t+2W(t)}$ (note that I wrote $-2t$ instead of $2t$, otherwise it does not hold, so I assume it is a typo) and given that ...

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The "most extreme" distributions would be ones containing $2N$ data points , half of which have a value of $a$ and the other half a value of $b$ The mean of such a distribution is $\mu = \frac12(a+b)$ the standard deviation is $\sigma = \frac12 |b-a|$ So for this distribution every point is exactly one standard deviation from the mean

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If you assume that the sequence has a normal distribution, with mean value $\mu$ and standard deviation $\sigma$, then the probability of getting a sample in the interval $\left[\mu-\sigma, \mu+\sigma\right]$ will be roughly equal to 68.27%. So when you have measured something, which has a normal distribution, for a finite time and calculate the mean and ...

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Comment: The very nice Answer by @snarfblatt gives both the derivation and the numerical answer. Sometimes this general type of problem is known as a 'random sum of random variables'. Here is a brief simulation in R, which may help you visualize the process. It is based on a million repetitions of your experiment. Answers should be accurate to two or ...

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The variance of a single dice roll is $\frac{35}{12}$. The outcomes of the dices are indepedent. Therefore the variance of the sum of n dice rolls is $n\cdot \frac{35}{12}$ You can consider each dice roll as a random variable. And for the sum of $n$ rolls you can apply the central limit theorem. For suffcient large number of dice rolls the sum of the ...

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Let $N$ take values in the natural numbers and $X_i$ be iid as the other variable; you want the variance of $Y=\sum_{i=1}^N X_i$. The conditional variance formula gives $$Var(Y)=Var(E(Y|N))+E(Var(Y|N))\\ =Var(NE(X_1))+E(NVar(X_1))\\ =(EX_1)^2Var(N)+Var(X_1)EN.$$ In case $N, X_i$ are taken from dice rolls, I get $EX_1=EN=\sum_{i=1}^6i/6=7/2,$ ...

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When you go for t statistic, it is assume that you do not know the population standard deviation. To answer your question, the sample standard deviation is calculated based on the below formula $$s = \sqrt{\frac{\sum_{1}^{n}(x_i -\bar x)^2}{n-1}}$$

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You are almost right except that there could be correlation between the portolio return ( devoid of exchange rate fluctuations) and the exchange returns. Let us say the variances are notated $\sigma_p^2$ and $\sigma_c^2$ and $\rho$ is the correlation coefficient. then the variance of the portfolio is given by \sigma_T^2 = \sigma_p^2 + \sigma_c^2 + ...

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If you let these numbers be $y$ values and allow corresponding $x$ values to be $1,2,3...$ and paired in sequence with the $y$ values, you could calculate the PMCC or the equation of the regression line.

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