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I have a homework question that I wasn't positive about. This is the first probability course I have taken and the class is only taught using excel so I apologize for the lack of formulas in my reasoning.

The size of a fish in a lake follows a normal distribution with a mean = 1 lb 4 oz and standard deviation = 3 oz. Fish that weigh less than 1 lb 9 oz must be released back into the lake. Bill wants to reach his limit of 5 'keepers'. What is the probability that Bill must catch at least 40 fish in order to reach the limit?

To find p I used norm.dist(25,20,3,1) = 0.190787. Then p(x=5) = binom.dist(35,40,p,0) = 0.09530982. Does this seem like the right way to do a problem like this? I'm not sure of how I would do it by hand as my professor does not teach us any formulas other than the excel ones. Any help is very much appreciated!!

user87654321
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2 Answers2

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What you want is the probability that after catching $39$ fish he will have at most $4$ keepers, meaning that he will have to catch at least $40$ fish in order to reach his limit. If I’m reading the Excel documentation correctly, you get this with BINOM.DIST(4,39,p,TRUE), where $p$ is the probability that a caught fish is a keeper: that returns the sum of the probabilities of having $0,1,2,3$, or $4$ keepers after $39$ catches. On my reading of the documentation the probability $p$ is given by the Excel function 1-NORM.DIST(25,20,3,TRUE), which ought to be about $0.04779$.

Brian M. Scott
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Define the following:

  • $X$ represents the weight of a single randomly caught fish.
  • $p$ represents the probability that a single randomly caught fish caught is a keeper.
  • $N$ is the total number of fish that are caught in order to obtain the $5^{\rm th}$ keeper.

The desired probability, then, is $\Pr[N \ge 40]$.

The value of $p$ is easy to find: it is simply $$p = \Pr[X \ge 25]$$ where $X \sim \operatorname{Normal}(\mu = 20, \sigma = 3)$, and all units of weight are expressed in ounces (1 pound = 16 ounces). Standardizing, we get $$p = \Pr\left[\frac{X - \mu}{\sigma} \ge \frac{25-20}{3}\right] = \Pr[Z \ge \tfrac{5}{3}] \approx 0.0477904.$$ The random variable $N$ is negative binomially distributed: specifically, $$\Pr[N = n] = \binom{n - 1}{n - r} p^r (1-p)^{n-r}, \quad n = r, r+1, r+2, \ldots,$$ where $r = 5$ is the number of keepers we need to observe, and $p$ is the probability of observing a keeper. Then $$\Pr[N \le 39] = \sum_{n=5}^{39} \Pr[N = n] \approx 0.0408402,$$ and the desired probability is $$\Pr[N \ge 40] \approx 0.962854.$$ Of course, this is also computed as a binomial probability, as mentioned by Brian M. Scott: $$\Pr[N \ge 40] = \Pr[Y = 4],$$ where $Y \sim \operatorname{Binomial}(n = 39, p)$, because the probability that at least $40$ fish need to be caught is equal to the probability that exactly $4$ keepers are caught among $39$ fish, requiring the fifth keeper to be at least the $40^{\rm th}$ fish.

heropup
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