Respuesta :
The probabilities for various binomial distribution have been determined.
The complete question is
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 285 employees. What is the probability that Company A will get the discount? (Round to four decimal places as needed.)
(b) Company B has 502 employees. What is the probability that Company B will get the discount? (Round to four decimal places as needed.)
(c) Company C has 1033 employees. What is the probability that Company C will get the discount? (Round to four decimal places as needed.)
What is Probability ?
Probability is a stream in mathematics that study the likeliness of an event to happen .
On the basis of the given data
(a) Let X is a random variable that denotes the number of employees that earn less than prevailing average.
Here X has binomial distribution with
n=285 and p=0.05.
As np and n(1-p) are greater than 5 so using normal approximation X has normal distribution with parameters
μ= np-285 * 0.05
= 14.25
standard deviation is given by
[tex]\sigma=\sqrt{np(1-p)}=\sqrt{285* 0.05* 0.95}\\\\=3.6793[/tex]
Applying continuity correction.
The z-score for X = 30-0.5 = 29.5 is
[tex]z=\dfrac{29.5-14.25}{3.6793}\\\\=4.14[/tex]
The probability that Company A will get the discount is given by
[tex]P(X\geq 30)=P(z > 4.14)=0.0000[/tex]
(b) Let X is a random variable that denotes the number of employees that earn less than prevailing average.Here X has binomial distribution with
n=502 and p=0.05.
[tex]\rm \mu=np=502* 0.05=25.1[/tex]
standard deviation
[tex]\rm \sigma=\sqrt{np(1-p)}=\sqrt{502\cdot 0.05\cdot 0.95}=4.8831[/tex]
Applying continuity correction.
The z-score for X = 30-0.5 = 29.5 is
[tex]\rm z=\dfrac{29.5-25.1}{4.8831}=0.90[/tex]
The probability that Company B will get the discount is
[tex]P(X\geq 30)=P(z > 0.90)=0.1841[/tex]
(c)Let X is a random variable that denotes the number of employees that earn less than prevailing average.Here X has binomial distribution with
n=1033 and p=0.05.
[tex]\mu=np=1033\cdot 0.05=51.65[/tex]
standard deviation
[tex]\rm \sigma=\sqrt{np(1-p)}=\sqrt{1033*0.05* 0.95}=7.0048[/tex]
Applying continuity correction.
The z-score for X = 30-0.5 = 29.5 is
[tex]\rm z=\dfrac{29.5-51.65}{7.0048}=-3.16[/tex]
The probability that Company C will get the discount is
[tex]\rm P(X\geq 30)=P(z > -3.16)=0.9992[/tex]
Therefore the probabilities for various binomial distribution have been determined.
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