Normally distributed salary increases

Question

Question

The annual percentage salary increases for workers are normally distributed with mean $12.2%$ and standard deviation $3.6%$. A random sample of $81$ of these workers was taken. What is the probability that more than half of the sample members had salary increases of less than $10%$?

Steps Taken:

Since the question asks for salary of less than $10%,$ I let $X$ be the salary increase.

$P(X\lt 0.1)=P(Z\lt (0.1-0.122)/(0.036))=P(Z\lt -0.611)=0.2709$

However, i do not know how to make use of the "more than half of the sample members" part of the question. Any help would be appreciated!

Answer 1

Let $X$ be the number of employees is the sample of $n = 81$ that got salary increases less than 10%. Then using what you have already found, you have $X \sim \mathsf{Binom}(n=81, p= 0.2709).$ You seek $P(X \ge 41) = 1 - P(X \le 40).$

Depending on the type and level of your course, you may be expected to find this probability using a normal approximation to the binomial distribution or to use software to find an exact probability.

In R statistical software the exact binomial probability is found to be very small.

1 - pbinom(40, 81, .2709)
[1] 5.859211e-06

Here is a plot of the PDF (or PMF) of $\mathsf{Binom}(n=81, p= 0.2709).$ The sum of the heights of the bars to the right of the broken red line is the probability you want.

The approximating normal distribution has $\mu = np = 80(.2709) = 21.9429$ and $\sigma = \sqrt{np(1-p)} = 4.0$ So $\mu \approx 22$ employees is many standard deviations away from 40 employees, and a normal approximation will also give a value very close to 0. The normal approximation from software is found below. You could also get a consistent result by standardizing and using normal tables; you would have to speculate about the exact result because the z-score will be off the table.

mu = 81*(.2709);  sg = sqrt(81*p*(1-p))
mu; sg
## 21.9429
## 3.999821
1 - pnorm(40.5, mu, sg)
## 1.746413e-06