Normal Approximation to Binomial Question

Question

I am having issues solving the following problem:

A recent study found that cedar trees by indigenous settlements grow taller than cedar trees not by indigenous settlements. The probability of a cedar tree being over 90m tall by an indigenous settlement is 0.20. If we take a random sample of 200 cedar trees growing near indigenous settlements, what is the probability that between 25 and 75 trees (exclusive) will be over 90m tall?

Now to start, I can see that N is large (200) and p is small (0.2), so I think this is a good candidate for normal approx. I can verify that np & nq > 5, which confirms this.

I need P(25 < X < 75) and add the continuity correction so

P(25.5 < X < 74.5) = P(X ≤ 74.5) - P(X ≤ 25.5)

μ = np = 40 and σ = 5.657 so to find z scores: (x-μ)/σ

P(Z ≤ 6.10) - P(Z ≤ -2.56)

The issue here is that 6 is a huge Z score! It is not on the table I am given, suggesting I have gone wrong somewhere? Could someone lend a hand?

Thanks!

Answer 1

It may be useful to compare the "Normal Approximation" approach with using the Binomial directly. Unless I'm mistaken, a quick computation using the Binomial shows this probability to be quite high (~0.9964).

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$X$: number of trees over 90m tall by an indigenous settlement

If we're ok with saying each tree in this sample is i.i.d., then $X \sim \text{Binomial}(n = 200, p = 0.20)$.

A quick calculation of $P(25 < X < 75)$ using MATLAB (MS Excel would be easy enough as well) shows this is quite high (~0.9964).

format long
pd = makedist('Binomial',200,0.20);
X = (26:74)';
>> sum(pdf(pd,X))
ans =
   0.996371237304574

@D... is correct.

>> normcdf(6.10,0,1)
ans =
   0.999999999469658

Answer 2

Your approach seems correct, although I didn't check the correctness of your computations. If $Z = 6.10$ is not in your table, this means $P(Z \leq 6.10) \approx 1$.