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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) $, but in this case $ n = y = 3 $ then $ P (Y = 3) = p ^ 3 $

$Z=\dfrac{X-μ}{σ}=\dfrac{X-120}{2}\Rightarrow p = (116

$P(-2

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

Thank you very much.

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    This is not a solution, or you should explain how you compute $p$. Note that the desired probability is $p^3$ anyway hence the détour by a binomial random variable is quite useless. – 2017-02-16
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    What is $p$? You need to get that from the normal distribution. – 2017-02-16
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    @Did You're right, it's not a solution. In fact, I want to know if my reasoning is right, then perform the calculations. Thank you for your contribution. – 2017-02-16
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    @BruceET $p$ is the probability of success, which I will calculate by leaving $X$ as the normal standard and then using tables I will calculate $P(116 – 2017-02-16
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    Please explain "using tables". – 2017-02-16
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    I mean [this](https://en.wikipedia.org/wiki/Standard_normal_table#Normal_and_standard_normal_distribution) – 2017-02-16
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    Did you ever get a solution? If you want to edit your problem to show what you have done, maybe someone can look at your methods and answers to see if it is going OK. – 2017-02-17
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    @BruceET There I realized the calculations that I thought as a solution. I hope they are understandable. – 2017-02-17
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    Right method, but seems table look-up may be incorrect. I verified in R statistical software: `p = diff(pnorm(c(116,118), 120, 2)); p; p^3` returns 0.1359051 for $p$ and 0.002510195 for $p^3.$ (Same as @A.G.) From my printed standard normal CDF table, $\Phi(-2) = 0.0228$ and $\Phi(-1) = 0.1587,$ so the difference is $p = 0.1587 - 0.0228 = 0.1359$ to four-place accuracy. If your table has only positive $z,$ then $\Phi(-1) = 1 - \Phi(1) = 1 - .8413 = 0.1587,$ by symmetry. Similarly for $\Phi(-2).$ Did you get your values on your own by using a table? Please let me know whether you resolve this. – 2017-02-17
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    I just found out: In my table the columns go from left to right: from $ 0.00 $ to $ 0.09 $ for the positives and from $ -0.09 $ to $ 0.00 $ for the negatives. I was just looking at that part in the digital version, which is why I compared it with other tables and did not match. Thank you very much to all. – 2017-02-17

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Your reasoning is correct, not your computation. You are indeed looking for $p^3$ where $$ p=P(116

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    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. – 2017-02-17
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    @emi: The usual definition of $\Phi$ is that $\Phi(z)=P(0$z$, $P(Z-z)=1/2-\Phi(-z)$. Therefore $P(Z<-1)=.5-\Phi(1)=.5-.341345=.158655$. Same for $-2$. – 2017-02-17