Respuesta :
We have been given that a normal distribution has a mean of 186.4 and a standard deviation of 48.9. We are asked to find the range of value that represents the upper 2.5% of the data.
We know that upper 2.5% of data would be 97.5% of data.
We will use z-score formula to solve our given problem.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
z = z-score,
x = Random sample score,
[tex]\mu[/tex] = Mean,
[tex]\sigma[/tex] = Standard deviation.
Now we will use normal distribution table to find z-score corresponding to 97.5% area or 0.975.
We can see from the normal distribution table that z-score corresponding to area 0.975 is [tex]1.96[/tex].
[tex]1.96=\frac{x-186.4}{48.9}[/tex]
Let us solve for x.
[tex]1.96\cdot 48.9=\frac{x-186.4}{48.9}\cdot 48.9[/tex]
[tex]95.844=x-186.4[/tex]
[tex]95.844+186.4=x-186.4+186.4[/tex]
[tex]282.244=x[/tex]
Therefore, the range [tex]x>282.244[/tex] represents the upper 2.5% of the data.
