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If the old data are $x_1$ to $x_N$, then the old variance is given by $$\sigma^2_{\text{old}}=\frac{1}{N}\sum_1^N x_i^2 -\overline{X}_{\text{old}}^2\tag{1}$$ and the new variance is given by $$\sigma^2_{\text{new}}=\frac{1}{N+1}\sum_1^{N+1} x_i^2 -\overline{X}_{\text{new}}^2.\tag{2}$$ You can see that $\sum_1^N x_i^2$ can be recovered from ...


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You can exploit the formulas $$\overline x=\frac1n\sum x_i=\frac{S_1}n,$$ $$\sigma^2=\frac1n\sum(x_i-\overline x)^2=\frac1n\sum x_i^2-\overline x^2=\frac{S_2}n-\overline x^2.$$ Every time you get a new value, update $n$, the sum of $x_i$ ($S_1$) and the sum of $x_i^2$ ($S_2$). From these you can compute the current mean, variance and standard deviation. ...


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We have $$ X \in N(111, 5) $$ The probability that a single match will last between 110 and 118 minutes is $$ p = P(110 \le X \le 118) = F_X(118) - F_X(110) $$ if $F_X$ is the cummulative dist. function. The probability that at least 30 (out of 64) lasts between these values is $$ 1 - \sum_{i=0}^{30} {64 \choose i}p^i(1-p)^{64-i} $$


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Find the probability $p$ that a single match will last between 110 and 118 minutes using the normal distribution. Then use the cumulative binomial distribution to find the probability of more than 30 succeeding.


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So this is not directly relevant to your final question, but it may help you understand a little bit as to why Standard Deviation/Variance are commonly used concepts. Variance is one of the moments of a probability distribution. I'm not sure what your math background is, so I won't go into what moments really are, but what's important to understand is that ...


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I assume the following distribution: $f(x)=\begin{cases} x, \ \ \text{if} \ \ 0 < x <1 \\ 2-x, \ \ \text{if} \ \ 1 \leq x <2 \\ 0, \ \ \text{elsewhere} \end{cases}$ $E(x)=\int_0^1 (x \cdot x) \ dx+\int_1^2 x \cdot (2-x) \ dx$. Now calculate E(x). greetings, calculus


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We assume that you want to find $E(X)$, and the variance of $X$. The expectation of $X$ is $\int_{-\infty}^\infty xf_X(x)\,dx$. In our case, this is $\int_0^1 (x)(x)\,dx+\int_1^2 (x)(2-x)\,dx$. We can calculate, but by symmetry of the density function (draw a picture) the answer is $1$. For the variance, we need to calculate $E(X^2)-(E(X))^2$. We have ...


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It depends on how your fifth variable depends on the other four. If it's the sum, then you sum the variance (by the way, this assumes that the four variables are independent of each other).



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