In a certain population, 15% of the people have Rh-negative blood. A blood bank serving this population receives 100 blood donors on a particular day.
So my teacher posted a solution to a problem below which solves the probability of getting 10 or less donors that are Rh-negative. I was wondering if I wanted to instead find the probability of 10 or MORE donors that are Rh-negative, how would I do so?
This was my attempt from following the teacher's logic
p(x≥10) = p(x>9.5) = 1 - p(Z>((9.5-17.1)/3.745)) = 1 - (Z>-2.03) = 1 - 0.2118 = 0.7882
Can someone confirm if I did this right, and if not, teach me how? I'm still a little lost.
$\underline{\textrm{Edit}}$:
Your flaw was that you´ve omitted a $0$: $P(Z>-2.03)=1-P(Z\color{blue}<-2.03)=1- 0.\color{red}02118=0.97882\approx 0.979$
Also note the blue inquality sign.
If you want to calculate a probability of a discrete random variable $X$ that is $x$ or more we have the following relation: $P(X\geq x)=1-P(X\leq x-1)$. Here we can apply the central limit theorem with $\mu=100\cdot 0.15=15$ and $\sigma=\sqrt{100\cdot 0.15\cdot 0.85}=3.57071$
In general if X is discrete we have $P(X\geq x)=1-P(X\leq x-1)\approx 1-\Phi\left( \frac{x-1+0.5-\mu}{\sigma}\right)$, where $0.5$ is the continuity correction factor. Inserting the values we obtain
$$P(X\geq 10)\approx 1-\Phi\left( \frac{10-1+0.5-15}{3.57071}\right)=1-\Phi\left( -1.54\right)=1-(1-\Phi\left( 1.54\right))=\Phi\left( 1.54\right)=0.93822\approx 93.8\%$$.
This is an approximation only. There are some thumb rules that allow us to apply the central limit theorem. One is that $\sigma^2>9\Rightarrow \sigma>3$. Since $\sigma=3.57071$ is not far away from $3$ we can suspect that the approximtion is not very good. The probability calculated with the binomial distribution is $94.49\%$.
Your attempt was almost right. $P(X\geq 10)=1-P\left( Z\leq \frac{10-1+0.5-17.1}{3.745}\right)=$ $1-P\left( Z\leq \frac{9.5-17.1}{3.745}\right)=1 -P(Z\leq -2.03)$.
Therefore $P(X\geq 10)\approx 1-\Phi\left(-2.03\right)=\Phi\left(2.03\right)$. If we use a table we can read off that $\Phi\left(2.03\right)=0.97882\approx 97.9\%$
Again this is not a very good approximation if we compare it with the (exact) result of BruceT which is $0.9907826\approx 99.1\%$. It depends on the purpose of the calculation if the approximated result is sufficiently accurate enough.