Pretend a person's water consumption follows a normal distribution. If on average he drinks 49 ounces of water in a day, and the variance of this behavior is 5 ounces-squared, what is his Z-score for a day when he drinks 57 ounces?

Question

Pretend a person's water consumption follows a normal distribution.  If on average he drinks 49 ounces of water in a day, and the variance of this behavior is 5 ounces-squared, what is his Z-score for a day when he drinks 57 ounces?

anonymous · Answer

Convert the variance to standard deviation by taking square-root, then plug all that in to the z-score formula.

anonymous · Answer

Thank you! I was confused because it said "ounces squared" so I was thinking that taking the square root would just cancel out the square...oops!

anonymous · Answer

Here's the formula in case you forgot it: http://stattrek.com/probability-distributions/standard-normal.aspx?Tutorial=Stat

anonymous · Answer

Taking the square-root does cancel the square, so you'll get σ=√5 oz. This is necessary to match the units of x and µ.