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.)
<h3>What is Probability ?</h3>
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
![\sigma=\sqrt{np(1-p)}=\sqrt{285* 0.05* 0.95}\\\\=3.6793](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%7Bnp%281-p%29%7D%3D%5Csqrt%7B285%2A%200.05%2A%200.95%7D%5C%5C%5C%5C%3D3.6793)
Applying continuity correction.
The z-score for X = 30-0.5 = 29.5 is
![z=\dfrac{29.5-14.25}{3.6793}\\\\=4.14](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B29.5-14.25%7D%7B3.6793%7D%5C%5C%5C%5C%3D4.14)
The probability that Company A will get the discount is given by
![P(X\geq 30)=P(z > 4.14)=0.0000](https://tex.z-dn.net/?f=P%28X%5Cgeq%2030%29%3DP%28z%20%3E%204.14%29%3D0.0000)
(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.
![\rm \mu=np=502* 0.05=25.1](https://tex.z-dn.net/?f=%5Crm%20%5Cmu%3Dnp%3D502%2A%200.05%3D25.1)
standard deviation
![\rm \sigma=\sqrt{np(1-p)}=\sqrt{502\cdot 0.05\cdot 0.95}=4.8831](https://tex.z-dn.net/?f=%5Crm%20%5Csigma%3D%5Csqrt%7Bnp%281-p%29%7D%3D%5Csqrt%7B502%5Ccdot%200.05%5Ccdot%200.95%7D%3D4.8831)
Applying continuity correction.
The z-score for X = 30-0.5 = 29.5 is
![\rm z=\dfrac{29.5-25.1}{4.8831}=0.90](https://tex.z-dn.net/?f=%5Crm%20z%3D%5Cdfrac%7B29.5-25.1%7D%7B4.8831%7D%3D0.90)
The probability that Company B will get the discount is
![P(X\geq 30)=P(z > 0.90)=0.1841](https://tex.z-dn.net/?f=P%28X%5Cgeq%2030%29%3DP%28z%20%3E%200.90%29%3D0.1841)
(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.
![\mu=np=1033\cdot 0.05=51.65](https://tex.z-dn.net/?f=%5Cmu%3Dnp%3D1033%5Ccdot%200.05%3D51.65)
standard deviation
![\rm \sigma=\sqrt{np(1-p)}=\sqrt{1033*0.05* 0.95}=7.0048](https://tex.z-dn.net/?f=%5Crm%20%5Csigma%3D%5Csqrt%7Bnp%281-p%29%7D%3D%5Csqrt%7B1033%2A0.05%2A%200.95%7D%3D7.0048)
Applying continuity correction.
The z-score for X = 30-0.5 = 29.5 is
![\rm z=\dfrac{29.5-51.65}{7.0048}=-3.16](https://tex.z-dn.net/?f=%5Crm%20z%3D%5Cdfrac%7B29.5-51.65%7D%7B7.0048%7D%3D-3.16)
The probability that Company C will get the discount is
![\rm P(X\geq 30)=P(z > -3.16)=0.9992](https://tex.z-dn.net/?f=%5Crm%20P%28X%5Cgeq%2030%29%3DP%28z%20%3E%20-3.16%29%3D0.9992)
Therefore the probabilities for various binomial distribution have been determined.
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