Normal Approximation Homework question Use normal approximation to estimate the probability of passing a true/false test of 50 questions if the minimum passing grade is 70% and all responses are random guesses.
I got that the average is equal to (50)(0.5) = 25
and the standard deviation is sqrt(npq) = sqrt(50 * 0.5 * 0.5)
So P(X > 70) = 1 - P(z < (35 - 25) / sqrt(50 * 0.5 * 0.5))
However, the answer that I got which is 0.02 is incorrect and Im not sure why.
 A: The exact answer is that under random guessing, the
number correct is $X \sim \mathsf{Binom}(50, .5).$
As you say, you seek $P(X \ge 35) = 1 - P(X \le 34) = 0.0033,$ as computed in R, where pbinom is a binomial CDF.
1 - pbinom(34, 50, .5)
[1] 0.003300224

It seems you are asked to use a normal approximation.
The approximating normal distribution has $\mu = np = 25$
and $\sigma = \sqrt{np(1-p)} = \sqrt{12.5} = 3.535534.$
Let $Y \sim \mathsf{Norm}(\mu = np,\sigma=\sqrt{npq}).$
Using R we can find $P(Y \le 34.5) = 1 - P(Y \le 34.5) = 0.00360.$  From R:
1 - pnorm(34.5, 25, 3.5355)
[1] 0.003604507

The figure below illustrates the normal approximation (red curve) to the binomial distribution (blue bars). The desired
probability lies to the right of the vertical
dotted line.

x = 0:50;  PDF=dbinom(x, 50, .5)
hdr="BINOM(50, .5) with Approx. Normal Density"
plot(x, PDF, type="h", lwd=2, col="blue", main=hdr)
 abline(h=0, col="green2")
 curve(dnorm(x, 25, sqrt(12)), add=T, lwd=2, col="red")
 abline(v=34.5, lwd=2, lty="dotted")

However, if you standardize and use printed tables of the
CDF of the standard normal distribution, some rounding is
inevitable; still, you can get an answer close the the previous one.
$$1-P(Y < 34.5)\\= 1-P\left(Z = \frac{Y-np}{\sqrt{np(1-p)}} < \frac{34.5 - 25}{3.5344} = 2.8679\right)\\ \approx
1-P(Z < 2.87) \approx 0.0021,$$
where $Z$ is standard normal.
(34.5 - 25)/(3.5344)
[1] 2.687868

1 - pnorm(2.87, 0, 1)
[1] 0.002052359

