For certain workers, the mean wage is $6.00/hr, with a standard deviation of $0.25. If a worker is chosen at random, what is the probability that the worker's wage is between $5.75 and $6.25? Assume a normal distribution of wages.

Answer :

Given the following information,

[tex]\begin{gathered} \mu=6 \\ \sigma=0.25 \\ x_1=5.75 \\ x_2=6.25 \end{gathered}[/tex]

Given the formula for the z-score below,

[tex]z=\frac{x-\mu}{\sigma}[/tex]

To find the z-score of the worker's wage for x₁

[tex]z=\frac{x_1-\mu}{\sigma}=\frac{5.75-6}{0.25}=\frac{-0.25}{0.25}=-1[/tex]

To find the z-score of the worker's wage for x₂,

[tex]z=\frac{x_2-\mu}{\sigma}=\frac{6.25-6}{0.25}=\frac{0.25}{0.25}=1[/tex]

By the empirical rule, 68-95-99.7% of the z-score lies within the normal distribution of the worker's wage between $5.75 and $6.25 hence, the probability is 0.68.