[Math] Confused between normal and binomial dist.

probability distributionsstatistics

The fish in a lake have weights that are normally distributed with a mean of 1.3 kg and
a standard deviation of 0.2 kg.

(b) John catches 6 fish. Calculate the probability that at least 4 of the fish weigh more than 1.4 kg.

In checking the marking scheme of the exam which contained this problem, I found out a mentioning to the binomial theorem. Here is a screen shot of weird answer for me (since it mentions a binomial dist. when the problem is talking about Normal dist.) and I wish if somebody explains to me what's going on with the solution to this part (part b) in specific
Image

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Best Answer

You can calculate the probability that (1) a fish weighs more than $1.4$kg, or (2) a fish weighs $1.4$kg or less, using the normal distribution

If fish weights are independent, you can then use the binomial distribution to work out the probability that (1) $4$, $5$ or $6$ fish weigh more than $1.4$kg or (2) $0$ or $1$ fish weigh $1.4$kg or less.

Note that the $\le$ in (b) should be $\lt$

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[Math] Tail of binomial Distribution (Confused)

It should be $P(X = k) = \dbinom{n}{k} p^k(1-p)^{n-k}$ and hence $P(X \leq k) = \sum_{r=0}^k\dbinom{n}{r} p^r(1-p)^{n-r}$.

[Math] Differences between Binomial and Normal Distribution Models

Exact binomial probability:

If I understand correctly, you have $X \sim Binom(n=1000, p=.05),$ and you want to know $$P(|X/1000 - .05| > .015) = 1 - P(35 \le X \le 65) = 0.0242,$$ computed in R statistical software as

1-(pbinom(65, 1000, .05)-pbinom(34, 1000, .05))
## 0.02423487

Normal approximation:

This could be approximated to about two places using $X \sim Norm(\mu=50,\, \sigma=6.892).$ With continuity correction, this is $1 - P(34.5 < X < 65.5) = 0.0245,$ computed as

1 - diff(pnorm(c(34.5, 65.5), 50, 6.892))
## 0.02451349

Simulation:

Also, as you suggest, this probability can be approximated (to two or three places) by the following simulation: One generates a million realizations of $X \sim Binom(1000, .05)$ and checks what proportion of them satisfy the condition $|X/1000 - .05| > .015.$ The answer is 0.0244. (In R, cond is a logical vector with elements TRUE and FALSE, and the mean of a logical vector is its proportion of TRUEs.)

x = rbinom(10^6, 1000, .05)
cond = (abs(x/1000 - .05) > .015)
mean(cond)
## 0.024429

The three results are about as near to each other as can be expected. I am not quite clear how this probability relates to the hypothesis test you are trying to do. Also, I understand that R is not exactly the same as MatLab. (R is excellent open source software available free of charge at www.r-project.org for Windows, Mac, and UNIX operating systems.)

I hope you can understand the three procedures and make any necessary adaptations for your specific homework problem.

In the plot below, the histogram shows the simulated distribution of $X,$ the curve is the approximating normal density function, the purple dots are exact binomial probabilities, and the vertical red lines mark the desired boundaries.

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