During the winter months, a convenience store sells hot chocolate. Sales follow a normal distribution of μ = 200 and σ = 210.
Compare the z-scores of a single-day value of 250 cups with a 15-day average value of 120 cups. Which value is farther from the mean?
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The first thing we should do with this kind of problem is list our variables!
- μ = overall population mean= 200
- σ = standard deviation
- x=single day's sales
- z-scores= how many standard deviations you are from the mean
We are going to have to evaluate this problem for two different situations in order to compare them. The first data point we are asked to solve for is the z-score for a single day where 250 cups of hot chocolate are sold.
The equation for this is
z= (x-μ)/ σ
Plugging in our data points here, we get
z=(250-200)/210=50/210=0.2381
Next we solve for the z score of an event where an average of 120 cups of cocoa are sold over 15 days. In this case, since it is not just a one day event, We have to solve for the standard error of the mean before we can use our z score formula.
The equation for standard error of the mean is
σ x=(σ /(sqrt(n)))=(210/(sqrt(15)))=54.2218
Now we can use the mean z score formula
z2=(x-μ)/σx=(120-200)/54.2218=-80/54.2218=-0.14754
You use the absolute value of both z scores
0.2381>0.14754, so it is more rare to have a day where only 120 cups of cocoa are sold!
