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 276 employees. what is the probability that company a will get the discount? (round to four decimal places as needed.) (b) company b has 511 employees. what is the probability that company b will get the discount? (round to four decimal places as needed.) (c) company c has 1036 employees. what is the probability that company c will get the discount? (round to four decimal places as needed.)

Explanation:

Step1: Identify the binomial distribution parameters

Let \(X\) be the number of employees earning minimum - wage or less. \(X\sim B(n,p)\), where \(p = 0.05\).

Step2: Use the normal approximation to the binomial

For \(n\) large (\(np\geq5\) and \(n(1 - p)\geq5\)), \(X\) can be approximated by \(Y\sim N(np,np(1 - p))\).
For \(n = 276\):
\(\mu=np=276\times0.05 = 13.8\)
\(\sigma=\sqrt{np(1 - p)}=\sqrt{276\times0.05\times(1 - 0.05)}=\sqrt{13.8\times0.95}\approx3.62\)
To find \(P(X\geq30)\), using the continuity correction \(P(X\geq30)\approx P(Y\geq29.5)\)
\(z=\frac{29.5 - 13.8}{3.62}=\frac{15.7}{3.62}\approx4.34\)
\(P(Z\geq4.34)\approx0\)

For \(n = 511\):
\(\mu=np=511\times0.05 = 25.55\)
\(\sigma=\sqrt{511\times0.05\times0.95}=\sqrt{25.55\times0.95}\approx4.92\)
Using continuity correction \(P(X\geq30)\approx P(Y\geq29.5)\)
\(z=\frac{29.5 - 25.55}{4.92}=\frac{3.95}{4.92}\approx0.80\)
\(P(Z\geq0.80)=1 - P(Z < 0.80)=1 - 0.7881=0.2119\)

For \(n = 1036\):
\(\mu=np=1036\times0.05 = 51.8\)
\(\sigma=\sqrt{1036\times0.05\times0.95}=\sqrt{51.8\times0.95}\approx7.03\)
Using continuity correction \(P(X\geq30)\approx P(Y\geq29.5)\)
\(z=\frac{29.5 - 51.8}{7.03}=\frac{- 22.3}{7.03}\approx - 3.17\)
\(P(Z\geq - 3.17)=1 - P(Z < - 3.17)=1 - 0.0008=0.9992\)

Answer:

(a) \(0.0000\)
(b) \(0.2119\)
(c) \(0.9992\)