Minimum wages, hours of employment, and child labor are regulated by the:
1 answer:
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Answer:
a) The probability that a randomly selected LU graduate will have a starting salary of at least $30,400 = P(x ≥ 30400) = 0.0968
b) The probability that a randomly selected LU graduate will have a salary of exactly $30,400 = 0.000021421
c) Percentage of students that will receive a tax break = 29.12%
d) Total Number of graduates this year = 3,000
Explanation:
This is a normal distribution problem with
Mean = μ = $20,000
Standard deviation = σ = $8,000
a) The probability that a randomly selected LU graduate will have a starting salary of at least $30,400 = P(x ≥ 30400)
We first normalize or standardize $30,400
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (30400 - 20000)/8000 = 1.30
The required probability
P(x ≥ 30400) = P(z ≥ 1.30)
We'll use data from the normal probability table for these probabilities
P(x ≥ 30400) = P(z ≥ 1.30) = 1 - P(z < 1.30)
= 1 - 0.90320
= 0.0968
b) The probability that a randomly selected LU graduate will have a salary of exactly $30,400
Here, we will use the normal distribution formula. The normal distribution formula is presented in the attached image
P(X = x) = f(x) = [1 ÷ σ√(2π)] × e^(-0.5z²)
x = $30,400
σ = $8,000
z = 1.30
P(X = 30400) = f(30400) = 0.000021421
c) Individuals with starting salaries of less than $15600 receive a low income tax break.
What percentage of the graduates will receive the tax break?
Required probability = P(x < 15600)
We first normalize or standardize $15,600
z = (x - μ)/σ = (15600 - 20000)/8000 = -0.55
The required probability
P(x < 15600) = P(z < -0.55)
We'll use data from the normal probability table for these probabilities
P(x < 15600) = P(z < -0.55)
= 0.29116 = 29.116% = 29.12%
d) If 189 of the recent graduates have salaries of at least $32240, how many students
graduated this year from this university?
We first find the percentage of LU graduates with salaries more than $32240
Required probability = P(x ≥ 32240)
We first normalize or standardize $32,240
z = (x - μ)/σ = (32240 - 20000)/8000 = 1.53
The required probability
P(x ≥ 32240) = P(z ≥ 1.53)
We'll use data from the normal probability table for these probabilities
P(x ≥ 32240) = P(z ≥ 1.53) = 1 - P(z < 1.53)
= 1 - 0.93699
= 0.06301 = 6.301%
So, 6.301% of the graduates this year = 189
Total Number of graduates this year = (189/0.06301) = 2999.5 = 3000 graduates this year.
Hope this Helps!!!
