Let $X, X$ ~ $N (120,4)$ be an independent measure, what is the probability that three measurements are equal, when measured three times?

Question

The voltage (in volts) of a given circuit is a random variable $ X $ that is normally distributed with the parameters $ μ = 120 $ and $ σ ^ 2 = 4 $

If three independent measurements are taken, what is the probability that the three measurements are between $ 116 $ and $ 118 $ volts?

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My idea is to first get a probability of success $ p $, which I will calculate by standardizing $ X $, and then finding the probability that $ X $ is between $ 116 $ and $ 118 $.

Since I need to count the number of measurements, each one with probability of success $ p $ and each measurement attempt is done independently, I would do it with another variable $ Y $ ~ $ B (3, p) $

The answer to the question would be $ P (Y = 3) $, this case $ n = y = 3 $ then, $ P (Y = 3) = p ^ 3 $

$Z=\dfrac{X-μ}{σ}=\dfrac{X-120}{2}\Rightarrow$

$p = (116<X<118) = P(\dfrac{116-120}{2}<Z<\dfrac{118-120}{2}) = P(-2<Z<-1) = $

$\Phi(-1) - \Phi(-2) = 0,13786 - 0,01831 = 0,11955 \Rightarrow p^3 = 0,001708633$

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Is the correct way I'm thinking the solution to the exercise?

Thank you very much.

Answer 1

\begin{align} P( 116 \leq X \leq 118) &= P \left( \frac{116-120}{\sqrt{4}} \leq \frac{X-120}{\sqrt{4}} \leq \frac{118-120}{\sqrt{4}} \right) \\ & = P( -2 \leq Z \leq -1 ) \\ &= \Phi(-1) - \Phi(-2) \\ &= 0.1586553 - 0.02275013 \\ & = 0.1359051 \end{align}

If the measurements are independent then the probability that all three fall into this interval is $0.1359051^3 = 0.002510194 $