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

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?

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, $P (Y = 3) = p ^ 3$

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

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

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

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

Thank you very much.

• Your approach is right, but your values of the CDF ($0.13786$ and $0.01831$) are wrong. You must be reading a table, and looking at the wrong end of it. You could just use the pnorm function in R. Commented Feb 17, 2017 at 18:27
• @Bridgeburners For $\Phi(-1)$ I am reading in row $-1.0$ column $0.0$ and for $\Phi(-2)$ I am reading in row $-2.0$ column $0.0$, which row and column should I read? Thank you. Commented Feb 17, 2017 at 19:02

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