Question 1063274: Good day! Please help me in my math homework.
The salaries of employees in a company have a mean of $5,000 and a standard deviation of $1,000. What is the probability that an employee selected at random will have a salary of (a) more than $5,000, (b) between $5,000 and $6,000, and (c) more than $7,000?
Would really appreciate your help. Thank you!
Found 2 solutions by Boreal, rothauserc:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! If the mean is 5000, and the sd 1000.
Given a normal distribution, which isn't stated, but I will assume.
half the people will have a salary more than $5000.
z=(x-mean)/sd
=(6000-5000)/1000=1, so the probability of being between $5000 and $6000 is the probability of z being between 0 and 1, which is 0.3413
More than 7000 is p(Z>2), which is 0.0228.
Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! For this problem we use the normal probability(P) distribution and its associated z-values to determine probabilities
:
z-value = (X - mean) / standard deviation
:
(a) P ( X > 5000 ) = 1 - P ( X < 5000 )
z-value = ( 5000 - 5000 ) / 1000 = 0
P ( X > 5000 ) = 1 - 0.50 = 0.50
:
(b) P ( 5000 < X < 6000 ) = P ( X < 6000 ) - P ( X < 5000 )
z-value = ( 6000 - 5000 ) / 1000 = 1
P ( 5000 < X < 6000 ) = 0.8413 - 0.5000 = 0.3413
:
(c) P ( X > 7000 ) = 1 - P ( X < 7000 )
z-value = ( 7000 - 5000 ) / 1000 = 2
P ( X > 7000 ) = 1 - 0.9772 = 0.0228
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