Central Limit Theorem Probability

Question

 

Suppose that the average male drinks 2 liters of water when active outdoors (with a

standard deviation of 0.7 liters). Consider these to be the true population parameters.

You are planning a full day nature trip for 50 men and you are bringing 110 liters of

water. What is the probability of running out of water? Show your work and explain.

Answer 1

If 1 man drinks 2L of water then 50 men would drink (on average) 100L of water. 0.7L standard deviation is equivalent to 35L for 50 men.

So if we assume a normal distribution with X=110 and μ=100, σ=35:

Z=(X-μ)/σ=(110-100)/35=10/35=0.29 approx.

This is equivalent to 0.6125, that is, the probability of using 110L of water or less is 61.25%.

Therefore there is a probability of 100-61.25=38.75% of running out of water.

You could also work it out by dividing 110L by 50=2.2L per man, μ=2, σ=0.7.

Z=(2.2-2.0)/0.7=0.2/0.7=0.29.