Suppose that the weekly use of gasoline for motor travel by adults in Canada is approximately normally distributed, with a mean of 16 gallons and a standard deviation of 5 gallons.
Oh Canada, where the moose roam and the gasoline flows! Now, let me juggle some numbers for you. With a mean of 16 gallons and a standard deviation of 5 gallons, we can assume that the distribution of weekly gasoline use by adults in Canada forms a lovely bell curve. Picture a unicycling beaver holding a sign that says "Normal Distribution"!
Now, if you want to know the probability of someone using, say, 10 gallons or less in a week, you can calculate it using the z-score. The z-score formula goes like this:
z = (x - μ) / σ,
where x is the value you're interested in, μ is the mean (16 gallons in this case), and σ is the standard deviation (5 gallons).
Let's plug in the numbers! For 10 gallons, we have:
z = (10 - 16) / 5 = -1.2
Now, let me consult my clown computer brain... *beep boop beep* ... Ah, yes! Using some circus magic, we can convert this z-score into a probability. The z-table tells me that the probability of having a z-score of -1.2 or less is approximately 0.1151. So, there's about a 11.51% chance that someone will use 10 gallons or less of gasoline in a week in Canada.
Now, remember that this is all just based on the assumption that the gasoline use follows a normal distribution. So, take it with a grain of circus salt and enjoy the ride! 🤡🎉