Question

About 5% of hourly paid workers in a regioner the prevailing minimum wage or less A grocery chain offers discount rates to companies that have at least 30 employees who earn the prevailing minimum wage or less Complete parts
through (c) below

(a) Company Ahas 281 employees What is the probability that Company A will get the discount?

(Round to four decimal places as needed)

Answer 1

Answer:

0

Step-by-step explanation:

Let X  to be a random variable that looks a binomial distribution which denoted the number of employees out of the 281 who earn the prevailing minimum wage or less

The sample size n = 281

The population parameter p = 5% = 0.05

Using normal approximation for the mean.

$\mu = np$

$\mu = 281\times 0.05$

$\mu = 14.05$

The standard deviation is:

$\sigma = \sqrt{np(1-p)}$

$\sigma = \sqrt{281 \times 0.05(1-0.05)}$

$\sigma = \sqrt{281 \times 0.05(0.95)}$

$\sigma = \sqrt{13.3475}$

$\sigma =3.6534$

By using continuity correction; the sample mean x is:

x = 30 - 0.5

x = 29.5

The z statistic test can now be as follows:

$Z = \dfrac{x-\mu}{\sigma}$

$Z = \dfrac{29.5-14.05}{3.6534}$

$Z = \dfrac{15.45}{3.6534}$

Z = 4.23

Thus, the probability that company A will get a discount is

P(X ≥ 30) = P(Z >4.23)

= 1 - P(Z < 4.23)

By using the Excel function for the z score 4.23 i.e. "=1 - NORMSDIST(4.23)" we get;

= 0.0000