Answer:
The answer is below
Step-by-step explanation:
Given that mean (μ) = $32500, standard deviation (σ) = $2500.
a) The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
For x < 36000:
[tex]z=\frac{36000-32500}{2500}=1.4[/tex]
From the normal distribution table, P(x < 36000) = P(z < 1.4) = 0.9192 = 91.92%
b) One standard deviation of mean = μ ± σ = (32500 ± 2500) = (30000, 35000)
Two standard deviation of mean = μ ± 2σ = (32500 ± 2*2500) = (27500, 37500)
Empirical rule states that 68% of data falls within one standard deviation from the mean, 95% falls within two standard deviation from the mean and 99.7% falls within one standard deviation from the mean.
Hence 95% of salaries is between $27,500 and $37,500.
c) 95% of salaries is between $27,500 and $37,500.
P(x < 27500) = (100% - 95%) / 2 = 2.5%
d) If the z score is less than -3 or greater than 3, it is considered an outlier.
For x < 42000:
[tex]z=\frac{42000-32500}{2500}=3.8[/tex]
Hence $42000 is an outlier