# Probability statistics - probability that 10 or more of the donors are Rh-negative?

In a certain population, 18% of the people have Rh-negative blood. A blood bank serving this population receives 95 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.

• First, pay attention that 18% $\ne$ 15%, and 95 $\ne$ 100. – user158565 Jul 29 '19 at 1:18
• p(x≥10) = p(x>9.5) $\approx$ p(Z>((9.5-17.1)/3.745)) = p(Z>-2.03) =0.9788 – user158565 Jul 29 '19 at 3:20

Let $$X \sim \mathsf{Binom}(n = 95, p = .18).$$ Then you seek $$P(X \ge 10) = 1 - P(X \le 9) = 0.9841.$$

Directly in R, where pbinom is a binomial CDF, we have

1 - pbinom(9, 95, .18)
[1] 0.9840775


Using a normal approximation with continuity correction, you should begin with $$P(X \ge 10) = P(X > 9.5) = \cdots .$$ You have the right idea, except I guess you got mixed up using normal tables. In particular, $$P(Z>-2.03) = 0.9788 \ne 0.2218.$$

1 - pnorm(-2.03)
[1] 0.9788217


Below is a plot of the relevant part of the PDF of $$\mathsf{Binom}(95,0.18)$$ along with the normal distribution that matches the binomial mean and standard deviation.

x = 0:35;  PDF = dbinom(x, 95, .18)
plot(x, PDF, type = "h", lwd=2)
abline(h=0, col="green2")
abline(v = 9.5, col="red", lty="dotted")
mu = 95*.18;  sg = sqrt(mu*(1-.18))
curve(dnorm(x, mu, sg), add=T, lwd=2, col="blue")