Question

A normal distribution has a mean of 186.4 and a standard deviation of 48.9.

Answer 1

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.  

$z=\frac{x-\mu}{\sigma}$, where,

z = z-score,

x = Random sample score,

$\mu$ = Mean,

$\sigma$ = 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 $1.96$.

$1.96=\frac{x-186.4}{48.9}$

Let us solve for x.

$1.96\cdot 48.9=\frac{x-186.4}{48.9}\cdot 48.9$

$95.844=x-186.4$

$95.844+186.4=x-186.4+186.4$

$282.244=x$

Therefore, the range $x>282.244$ represents the upper 2.5% of the data.