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Sagot :
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.
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