probability/Confidence intervals

Question

The manufacturer of a coffee dispensing machine claims that the ounces per cup
has a distribution with mean μ = 0.7 ounces.  and standard deviation σ = 0.7 ounces. If 39 cups
are measured, what is the probability that the average ounces per cup
x bar is more than 8.2 ounces?

Answer 1

Jaspreet,

 

First, you have a typo on the mean value which placed the decimal point wrong. With your 0.7 oz value and a Normal distribution, 17% of the customers would get no coffee, and no customer would ever get more than 3.0 oz of coffee.  Bad machine!  So the correct mean value is probably 7.0 oz.  Then over 95% of the customers would get between 5 and 9 oz of coffee, still a shoddy machine because its standard deviation is so large relative to the mean.

 

Second, why is the Normal distribution only an approximation? Two basic reasons. The theoretical Normal distribution allows for "negative values" - not possible when you're pouring coffee - and it also allows very large values that would overflow any cup size. Admittedly, both cases can only occur with very, very small probabilities. So they can be ignored. In practice we only care about the range within ±3σ, not the theoretical tails.  The other reason is that when a final random variable (amount of coffee poured) is the result of many other contributing factors in the machine (which don't have to be Normally distributed), then the final random variable is still approximately Normally distributed.

 

The final reason we use the Normal distribution is that we have a formula for it and tabular data that can be used to solve  problems.

 

The N(0,1) z-variable for your data is z = (x - μ)/σ = (8.2 - 7.0)/0.7 = 1.2/0.7 = 1.714

And the probability of exceeding this z-value is  0.0432

 

Times 39 cup sample = 1.68 cups.  So you may see one or two cups in your sample of 39 cups which exceed 8.2 oz.