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Disclaimer: I'm not an expert, so please correct me if there are serious flaws in my argument.
Hand-waving/guessing
First observe that if $X^2=t$ then $X$ is either $\sqrt{t}$ or $-\sqrt{t}$, so this is a discrete distribution, and to specify the distribution you only need to compute $P(X=\sqrt{t} \mid X^2=t)$ and $P(X=-\sqrt{t} \mid X^2=t)$.
In analogy with ...
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It seems the task is to find $P(X > 2) = 1-P(X \le 2)$ when $X \sim \mathsf{Binom}(n = 100, p=0.02).$ So the computation reduces to finding $P(X \le 2) = 0.6767,$ to four places. Exact computation in R is as below: first using the R binomial CDF function pbinom; second, using R as a calculator to compute the necessary three terms of the binomial PDF ...
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I'm wondering if this conditional distribution is even well$-$defined. Let $\epsilon_1,\epsilon_2$ be very small positive numbers and $t>0$ be arbitrary. Define $I_1=(-\sqrt{t}-\epsilon_1,-\sqrt{t}+\epsilon_1)$ and $I_2=(\sqrt{t}-\epsilon_2,\sqrt{t}+\epsilon_2)$. Assume $\epsilon_1, \epsilon_2$ are small enough to make $I_1,I_2$ disjoint. Notice how $$P\...
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MAJOR EDIT
Better check @angryavian's answer.
$\Pr\{X=x|X^2=t\}$
$\;\;\;=\Pr\{X=x|X>0, X^2=t\}\Pr\{X>0|X^2=t\}$
$\;\;\;\;\;\; +\Pr\{X=x|X<0, X^2=t\}\Pr\{X<0|X^2=t\}$
Now:
$\Pr\{X=x|X>0, X^2=t\}=1$, if $x=\sqrt{t}$, and zero otherwise,
also $\Pr\{X=x|X<0, X^2=t\}=1$, if $x=-\sqrt{t}$, and zero otherwise.
And:
$\Pr\{X>0|X^2=t\} = \dfrac{\...
2
Determinant of $\Sigma$ may be zero.
Let $k$ be the dimension of the space to which the sample vectors $x_n$ belong. If $N < k$ then $\Sigma$ is not full rank and therefore $\det \Sigma = 0$.
Moreover, even when $N \ge k$ it is still possible that $\det \Sigma = 0$. For a simple example, suppose that $k=2$ and that the sample vectors $x_n \in\mathbb{R}^2$ ...
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Let's start form
$$\frac{X}{\sigma}\sim N(0;1)$$
It is well known, and anyway easy to prove that
$$W=\Bigg(\frac{X}{\sigma}\Bigg)^2\sim \chi_{(1)}^2=Gamma(1/2;1/2)$$
Thus
$$Y=\sigma^2\cdot W=Gamma(1/2;1/(2\sigma^2))$$
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Hint:
Write your function as a difference of CDF evaluating at two different points. Then differentiate.
Also, you should know that $F'(z)=f(z)$ where $F$ is the CDF and $f$ the PDF.
Notes.
$$
P(-\sqrt{z}\le X\le \sqrt{z})=P(X\le \sqrt{z})-P(X\le -\sqrt{z})=F(\sqrt{z})-F(-\sqrt{z})
$$
Now use the chain rule.
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Could you please elaborate last step? How did $\sigma^2$ fit in distribution?
Let's have
$$X\sim Gamma(n;\theta)$$
(I mean $\theta$ is the rate parameter)
$$f_X(x)=\frac{\theta^n}{\Gamma(n)}x^{n-1}e^{-\theta x}$$
Let's set
$$y=\sigma^2 x$$
$$x=\frac{y}{\sigma^2}$$
$$x'=\frac{1}{\sigma^2}$$
thus
$$f_Y(y)=f_X[g^{-1}(y)]\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|=\frac{\...
1
In normal distribution we use the notation $P(X\le a)=\phi(a)$
$P\bigg(\dfrac{x-\mu}{\sigma} < \dfrac {25-\mu}{\sigma}\bigg)=P\bigg(Z< \dfrac {25-\mu}{\sigma}\bigg)=\phi\bigg(\dfrac{25-\mu}{\sigma}\bigg)=0.1082$
Now look at normal table and find $a$ such that $\phi(a)=0.1082$ then,
$a=\dfrac{25-\mu}{\sigma}$
similarly, you have $P(X>42)=.1303$
So ...
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Let $X = y + \sigma Y$, where $Y \sim N(0,1)$.
We know that $k = \mathbb P(X<x) = \mathbb P(Y < \frac{x-y}\sigma)$, therefore
$$\mathbb P(X < 2y-x) = \mathbb P(Y < \frac{y-x}\sigma) = 1- \mathbb P(Y \geq \frac{y-x}\sigma) = 1 - \mathbb P(Y \leq \frac{x-y}\sigma) = 1 - k$$
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The point is that you calcluated the probability that Ali makes exactly one jump over $6$m. It is possible that Ali cleared $6$m on two, or even all three, jumps. So you need to take those possibilities into account too.
The easiest way to do this: Ali fails to make the team if and only if he jumps less than $6$m on all three attempts. This occurs with ...
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