I have this exercise:
We learned in Exercise 3.26 that about 90% of American adults had chickenpox before adulthood. We now consider a random sample of 120 American adults.
(a) What is the mean and standard deviation.
(c) What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood?
I have used Z score to calculate it, since both $np$ and $n(1-p)$ is above 10, we can use normal distribution. First I calculated mean and SD from exercise (a) for a binominal distribution, then the Z score:
Mean:
$μ=np$
$μ=120*0.90$
$μ=108$
Then the Standard Deviation:
$σ=\sqrt{np(1-p)}$
$σ=\sqrt{120*0.90(1-0.90)}$
$σ=\sqrt{10.8}$
$σ=3.28$
Now I calculate the Z-score:
$Z=(observation-mean)/SD$
$Z= (105-108)/3.28$
$Z=-0.91$
We will look up that number on the Z-table: 0.1814 = 18.14%
Is this correctly done? I tried looking it up on the internet, but I see people getting 21% when they use pbinom(105,120,.9) from the R language = 0.2181634 = 21%
The same website calculates it this way, without R and still gets 21%
Best Answer
I think you've done the steps correctly, the discrepancy between the binomial and normal approximation seems believable in this case (if your $p$ is quite far from $0.5$ and your $n$ is not enormously large, then your answer can usually be a few percentage points off).
To get a more accurate answer using the normal approximation, some people advocate using a continuity correction. This means calculating the $z$-score as $\left(105.5 - 108\right)/3.28$ instead of $\left(105 - 108\right)/3.28$. Using this continuity correction gives an answer of $22.3\%$ which is a bit closer to the true probability compared to your answer.
Ultimately there are a few ways to go about solving this problem, some by using approximations which lead to slightly different answers. The steps you've taken appear sound, assuming your course has taught you to solve it that way.