In a company there are 7000 employees and 2000 of them have been with the compan
ID: 3228916 • Letter: I
Question
In a company there are 7000 employees and 2000 of them have been with the company for less than 3 years The salaries of the employees are roughly distributed according to a Gaussian distribution with mean $60,000 and variance ($10,000)^2 for employees who have been with the company for less than 3 years and are roughly Gaussian with mean $80,000 and variance ($20,000)^2 if the employees have been with the company for 3 years or longer. If you know that Mr. Smith's salary is between $60,000 and $70,000 what is the probability that Mr. Smith has been with the company for less than 3 years? If you know that Mr. White's salary is between $65,000 and $75,000 what is the probability that Mr. White has been with the company for less than 3 years? If you know that Mr. Green's salary is between $65,000 and $75,000 what is the probability that Mr. Green has been with the company for less than 3 years?Explanation / Answer
(a) Mr. Smith salary is between $60000 to $70000, so
Probability that Mr. smith has been with the company for less than 3 years = Pr( Mr. smith less than 3 years) = P(X) let assume
P(X) = Pr ( salary of mr. smith between 60 K to 70K when working for less than 3years) * Pr ( working for less then three years) / [ Pr ( salary of mr. smith between 60 K to 70K when working for less than 3years)*Pr ( working for less then three years) + Pr ( salary of mr. smith between 60 K to 70K when working for more than 3years)* Pr ( working for more then three years) ]
Pr ( salary of mr. smith between 60 K to 70K when working for less than 3years) = here mean salary is equal to $ 60k and standard deviation equals to $10 k.
so Pr ( 60 k < salary < 70k ; 60k ; 10k ) by normal distribution = (Z = (70 - 60)/60) - (0) = (1) - (0)
where is normal probability distribution function
Pr ( 60 k < salary < 70k ; 60k ; 10k ) = 0.8413 - 0.5 = 0.3413
Now, we will calculate
Pr ( salary of mr. smith between 60 K to 70K when working for more than 3years)] here mean salary is equal to $ 80k and standard deviation equals to $20 k.
so Pr ( 60 k < salary < 70k ; 80k ; 20k ) by normal distribution = (Z = (70 - 80)/20) - (Z = (60 - 80)/20) = (-0.5) - (-1)
where is normal probability distribution function
Pr ( 60 k < salary < 70k ; 80k ; 20k ) = 0.3085 - 0.1587= 0.1498 = 0.15
so Pr( Mr. smith less than 3 years) = 0.3413 * 2/7 / (0.3413 * 2/7+ 0.1498 * 5/7) = 0.4768
(ii)
Mr. White salary is between $65k to $75k, so
Probability that Mr. White has been with the company for less than 3 years = Pr( Mr. white less than 3 years) = P(Y) let assume
P(X) = Pr ( salary of mr. white between 65 K to 75K when working for less than 3years) * Pr ( working for less then three years) / [ Pr ( salary of mr. white between 65 K to 75K when working for less than 3years)*Pr ( working for less then three years) + Pr ( salary of mr. white between 65 K to 75K when working for more than 3years)* Pr ( working for more then three years) ]
Pr ( salary of mr. whitebetween 65 K to 75 K when working for less than 3years) = here mean salary is equal to $ 60k and standard deviation equals to $10 k.
so Pr ( 65 k < salary < 75k ; 60k ; 10k ) by normal distribution = (Z = (75 - 60)/60) - (Z = (65-60)/10) = (1.5) - (0.5)
where is normal probability distribution function
Pr ( 60 k < salary < 70k ; 60k ; 10k ) = (1.5) - (0.5) = 0.9332 - 0.6915 = 0.2417
Now, we will calculate
Pr ( salary of mr. white between 65 K to 75K when working for more than 3years)] here mean salary is equal to $ 80k and standard deviation equals to $20 k.
so Pr ( 65 k < salary < 75k ; 80k ; 20k ) by normal distribution = (Z = (75-80)/20 ) - (Z = (65-80)/20 ) = (-0.25 ) - (-0.75)
where is normal probability distribution function
Pr ( 65 k < salary < 75k ; 80k ; 20k ) = (-0.25 ) - (-0.75) = 0.4013 - 0.2266= 0.1747
so Pr( Mr. smith less than 3 years) = 0.2417 * 2/7 / (0.2417 * 2/7+ 0.1747 * 5/7) = 0.3563
(c) This part is also same as part b but only name changed here so
Pr(Mr green has been working for less than 3 years) = 0.3563