Question

The amount of water person typically drinks in a week has a mean of 100 ounces and a standard deviation of 15 ounces. Using the sd rule, what amount of water represents the top 97.5%? That is, what is the 97.5th percentile or the amount that only 2.5% are greater than?

Answer 1

Solution: We are given:

$\mu=100,\sigma=15$

We have to find the amount of water that represents the top 97.5%.

We need to find the z value corresponding to probability to 0.975. Using the standard normal table, we have:

$z(0.975)=1.96$

Now using the z score formula, we have:

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

$1.96=\frac{x-100}{15}$

$1.96 \times 15 = x-100$

$29.4=x-100$

$x=100+29.4 = 129.4$

Therefore, 129.4 ounces amount of water represents the top 97.5%