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)

Respuesta :

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.

[tex]\mu = np[/tex]

[tex]\mu = 281\times 0.05[/tex]

[tex]\mu = 14.05[/tex]

The standard deviation is:

[tex]\sigma = \sqrt{np(1-p)}[/tex]

[tex]\sigma = \sqrt{281 \times 0.05(1-0.05)}[/tex]

[tex]\sigma = \sqrt{281 \times 0.05(0.95)}[/tex]

[tex]\sigma = \sqrt{13.3475}[/tex]

[tex]\sigma =3.6534[/tex]

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:

[tex]Z = \dfrac{x-\mu}{\sigma}[/tex]

[tex]Z = \dfrac{29.5-14.05}{3.6534}[/tex]

[tex]Z = \dfrac{15.45}{3.6534}[/tex]

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

The discount value is "0".

Let X be a random variable that shows how many employees out of 281 earn the prevailing minimum wage or less.

X does have a binomial distribution with parameters [tex]n = 281 \ \ and\ \ p = 0.05[/tex]:

Since [tex]np \ \ and\ \ n(1-p)[/tex] are greater than 5, X does have a normal distribution of parameters when using the normal approximation.

[tex]\to \mu= np=281 \times 0.05= 14.05[/tex]

Calculating the standard deviation:

[tex]\to \sigma=\sqrt{np(1-p)}\\\\[/tex]

       [tex]= \sqrt{281 \times 0.05 (1-0.05)}\\\\=\sqrt{281 \times 0.05 \times 0.95}\\\\ =\sqrt{14.05 \times 0.95}\\\\ =\sqrt{13.3475}\\\\=3.653[/tex]

Continuity correction is required in this case. The z-score for [tex]X = 30-0.5 = 29.5[/tex]:

[tex]\to z= \frac{X- \mu}{\sigma}[/tex]

      [tex]=\frac{29.5-14.05}{3.653}\\\\=\frac{15.45}{3.653}\\\\=4.2294\\\\[/tex]

The likelihood that Company Its will receive the discount:

[tex]\to P(X\geq 30)=P(z>4.229)=0.0000117\approx \ 0.0000[/tex]

Therefore, the final answer is "0".

Learn more about the discount:

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