Respuesta :
The variable of interest is X: earnings of one worker.
This variable has a normal distribution with mean μ=$7.00/hr and standard deviation σ=$0.25
You need to determine the probability that a worker chosen at random earns between $6.50 and $7.50, symbolically:
[tex]P(6.50\leq X\leq7.50)[/tex]To determine the probability within these two values you have to subtract the accumulated probability until 6.50 to the accumulated probability until 7.50:
[tex]P(6.50\leq X\leq7.50)=P(X\leq7.50)-P(X\leq6.50)[/tex]First, you have to determine the corresponding values of the variable under the standard normal distribution, i.e., you have to calculate the Z-values for 7.50 and 6.50:
[tex]Z=\frac{X-\mu}{\sigma}N\sim(0,1)[/tex][tex]P(X\leq7.50)=P(Z\leq\frac{7.50-7}{0.25})=P(Z\leq2)[/tex][tex]P(X\leq6.50)=P(Z\leq\frac{6.50-7}{0.25})=P(Z\leq-2)[/tex]Next, you have to look for the accumulated probability under the standard normal distribution:
[tex]P(Z\leq2)=0.977[/tex]and
[tex]P(Z\leq-2)=0.023[/tex]Now that you have determined the probabilities you can determine the asked one:
[tex]\begin{gathered} P(6.5\leq X\leq7.5)=P(X\leq7.5)-P(X\leq6.5)= \\ P(Z\leq2)-P(Z\leq-2)=0.977-0.023=0.954 \end{gathered}[/tex]There is a 0.954 or 95.4% of probability that the worker choosen at random earns between $6.50/hour and $7.50/hour.
