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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?

Respuesta :

Solution: We are given:

[tex]\mu=100,\sigma=15[/tex]

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:

[tex]z(0.975)=1.96[/tex]

Now using the z score formula, we have:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

[tex]1.96=\frac{x-100}{15}[/tex]

[tex]1.96 \times 15 = x-100[/tex]

[tex]29.4=x-100[/tex]

[tex]x=100+29.4 = 129.4[/tex]

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

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