Question

about 5% of hourly paid workers in a region earn 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 (a) through (c) below. (a) company a has 257 employees. what is the probability that company a will get the discount? (round to four decimal places as needed.) (b) company b has 523 employees. what is the probability that company b will get the discount? (round to four decimal places as needed.) (c) company c has 1013 employees. what is the probability that company c will get the discount? (round to four decimal places as needed.)

Explanation:

Step1: Identify the binomial - normal approximation conditions

Let \(X\) be the number of employees who earn the minimum wage or less. \(X\sim B(n,p)\) where \(p = 0.05\). We use the normal approximation \(X\sim N(np,np(1 - p))\) when \(np\geq5\) and \(n(1 - p)\geq5\).
For \(n = 257\), \(np=257\times0.05 = 12.85\geq5\) and \(n(1 - p)=257\times(1 - 0.05)=257\times0.95 = 244.15\geq5\).
For \(n = 523\), \(np = 523\times0.05=26.15\geq5\) and \(n(1 - p)=523\times0.95 = 496.85\geq5\).
For \(n = 1013\), \(np=1013\times0.05 = 50.65\geq5\) and \(n(1 - p)=1013\times0.95=962.35\geq5\).
The continuity correction: To find \(P(X\geq30)\), we find \(P(X > 29.5)\) when using the normal approximation.

Step2: Calculate the mean and standard deviation

The mean \(\mu=np\) and the standard deviation \(\sigma=\sqrt{np(1 - p)}\).

• For Company A (\(n = 257\)):

\(\mu_{A}=np=257\times0.05 = 12.85\), \(\sigma_{A}=\sqrt{257\times0.05\times(1 - 0.05)}=\sqrt{12.85\times0.95}\approx\sqrt{12.2075}\approx3.494\)
\(z=\frac{x-\mu}{\sigma}\), so \(z=\frac{29.5 - 12.85}{3.494}=\frac{16.65}{3.494}\approx4.765\)
\(P(X\geq30)=P(Z\geq4.765)\approx0\) (using the standard - normal table \(P(Z\geq3.5)\approx0.0002\), and for \(z = 4.765\) the probability is extremely small)

• For Company B (\(n = 523\)):

\(\mu_{B}=np = 523\times0.05=26.15\), \(\sigma_{B}=\sqrt{523\times0.05\times(1 - 0.05)}=\sqrt{26.15\times0.95}\approx\sqrt{24.8425}\approx4.984\)
\(z=\frac{29.5 - 26.15}{4.984}=\frac{3.35}{4.984}\approx0.672\)
\(P(X\geq30)=P(Z\geq0.672)=1 - P(Z < 0.672)\)
From the standard - normal table \(P(Z < 0.672)\approx0.749\), so \(P(Z\geq0.672)=1 - 0.749 = 0.251\)

• For Company C (\(n = 1013\)):

\(\mu_{C}=np=1013\times0.05 = 50.65\), \(\sigma_{C}=\sqrt{1013\times0.05\times(1 - 0.05)}=\sqrt{50.65\times0.95}\approx\sqrt{48.1175}\approx6.937\)
\(z=\frac{29.5 - 50.65}{6.937}=\frac{- 21.15}{6.937}\approx - 3.05\)
\(P(X\geq30)=P(Z\geq - 3.05)=1 - P(Z < - 3.05)\)
From the standard - normal table \(P(Z < - 3.05)=0.0011\), so \(P(Z\geq - 3.05)=1 - 0.0011 = 0.9989\)

Answer:

(a) \(0.0000\)
(b) \(0.2510\)
(c) \(0.9989\)