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There are lots of salmon in a pond and their length (in centimeters) obeys normal distribution $N(70, 5.4^2)$. You and your friend go fishing and decide to continue fishing until both of you catch at least one salmon. What is the probability that your first salmon is $10$ cm longer than your friend's first salmon?

Right answer: $0.10$

I started doing it like this:

$$\begin{align} &E(X) = 70\\ &\sigma = 5.4\\ &Z = \frac{X2 - X1 - E(X)}{D}\\ &P(X2 - X1 \geq 10) = 1 - P(X2 - X1 \leq 10)\\ &= 1 - P(Z \leq (10 - 70)/5.4 ) = 1 - \Phi(-11.11) = 1 - (1 - \Phi(11.11)) = \Phi(11.11) \end{align} $$

Obviously this is wrong because even if I knew the value of $\Phi(11.11)$ it would be something like $0.999\ldots$ and not $0.10$ like it should be.

Any ideas what is the right way to do this?

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2 Answers 2

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The normal distribution has the property that the distribution of a sum of independent (not necessarily identically distributed) normal random variables is itself normal with the following properties: $$ \begin{align} \textrm{If}\;&X_i \sim N(\mu_i, \sigma^2_i)\\ \textrm{Then}\;&Y = \sum{a_iX_i} \sim N(\sum a_i\mu_i, \sum a_i^2\sigma^2_i) \end{align} $$ (DeGroot 1989, p.270)

Approaching your problem, what we want is the distribution of the random variable $Y = X_1 - X_2$ where $X_1$ is the length of your fish and $X_2$ is the length of your friend's fish. Given the data, $Y$ should be normally distributed as $Y \sim N\left((70 +(-1\cdot70)), (1^25.4^2 + -1^25.4^2)\right)$ or $Y \sim N(0, 58.32)$.

The probability of having $Y \geq 10$ is then: $$ P(Y \geq 10) = P\left(Z \geq \frac{10 - 0}{\sqrt{58.32}}\right) \approx 1-\Phi(1.309457) \approx 0.09519\;(9.519\%) $$

So either the $10\%$ is an approximation or I've made an error somewhere :)

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    $\begingroup$ Thank you very much for help. I had been struggling with this because I didn't know that you need to sum the expectation values and variances like that. One correction to your answer though: you forgot to mark the 1- before Φ(1.309457). Other than that it's perfect :) $\endgroup$
    – JZ555
    Commented Sep 11, 2013 at 8:32
  • $\begingroup$ My pleasure, and thanks for the correction; good catch! Also, summing the variance and means is a property of the Normal. Some distributions have similar closed-form sum formulæ under certain conditions; some don't have any nice form. $\endgroup$
    – Avraham
    Commented Sep 11, 2013 at 8:33
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The trouble is that $E(X)=70$ is not the expectation of $X_2-X_1$ hence your $Z$ has nothing to do with the centering needed to use the standard normal CDF $\Phi$.

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