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Suppose that a point(X, Y )is to be chosen at random in thexy-plane, whereXandYare independent random variables, and each has the standard normal distribution. If a circle is drawn in thexy-plane with its center at the origin, what is the radius of the smallest circle that can be chosen for there to be a probability of 0.99 that the point(X, Y )will lie inside the circle?

Short Answer

Expert verified

The radius of the smallest circle is 9.210

Step by step solution

01

Given information

XandYare independent random variables, and each has the standard normal distribution

02

Calculating the radius

Let r denote the radius of the circle. The point (X, Y) will lie inside the circle if and only if \({X^2} + {Y^2} < {r^2}\) . It also \({X^2} + {Y^2}\)has \({\chi ^2}\) a distribution with two degrees of freedom.

The smallest circle that can be chosen for there to be a probability of 0.99

\(\begin{align}P\left( {{X^2} + {Y^2} < {r^2}} \right) &= 0.99\\p\left( {{\chi ^2}(2) < {r^2}} \right) &= 0.99\\{r^2} &= 9.210\end{align}\)

Therefore, they must have \({r^2} \ge 9.210\)

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