Answer:
a) 60 years
b) 97.72th percentile
c) 52.65 years
Explanation:
Given that:
mean (μ) = 60 years and standard deviation (σ) = 2.5. The z score is given by the equation:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
a) the median correspond to 50th percentile.
The 50th percentile from the z table corresponds to a z score of 0. Using the z score equation:
[tex]z=\frac{x-\mu}{\sigma}\\0=\frac{x-60}{2.5}\\ x-60 = 0\\x=60[/tex]
The median retirement age is 60 years
b) When age is 65 years.
[tex]z=\frac{x-\mu}{\sigma}=\frac{65-60}{2.5} =2[/tex]
From the normal distribution table: P(x < 65) = P(z < 2) = 0.9772 = 97.72th percentile
c) The 16th percentile from the z table corresponds to a z score of -2.94. Using the z score equation:
[tex]z=\frac{x-\mu}{\sigma}\\-2.94=\frac{x-60}{2.5}\\ x-60 = -7.35\\x=60-7.35=52.65[/tex]
An age of 52.65 corresponds to 16th percentile