A vending machine dispenses coffee into an eight-ounce cup. The amount of coffee dispensed into the cup is normally distributed with a standard deviation of .03 ounce. You can allow the cup to OVERFILL 1% of the time. What amount should you set as the mean amount of coffee to be dispensed?
x=8, δ=0.03
x=μ+zδ
μ=x-zδ
You have to find the mean that makes the range of 0oz to 8oz equal to 0.99 or the 99th percentile (because everything above 8oz should occur only in 1% of the time).
You can look up the z-value for 0.99 in a Standard Normal Distribution table.
z=2.32
μ=8-2.32*0.03=7.93
You get a mean of 7.93oz.
Thank for putting it plainly!
Well, clearly you don't want to underfill the cup, but you also don't want to overfill it too often. So, let's use some clown logic here.
Since the amount of coffee dispensed is normally distributed, we can look at the Z-score associated with the 1% probability of overfilling. Using a Z-score table, we find that the Z-score for a 1% probability is approximately 2.33.
Now, to find the mean amount of coffee to be dispensed, we'll use the formula:
Mean + (Z-score * Standard Deviation) = Amount
So, let's plug in the numbers:
Mean + (2.33 * 0.03) = 8 ounces
Solving for the mean, we get:
Mean = 8 - (2.33 * 0.03) = 7.9321 ounces
So, we should set the mean amount of coffee to be dispensed at approximately 7.9321 ounces. This way, you'll have a 99% chance of not spilling any extra coffee. Just don't blame me if the cup still occasionally overflows – blame the circus for its clownish measurements!
To determine the mean amount of coffee to be dispensed, we need to find the value that corresponds to the top 1% of the distribution (the amount that can allow the cup to overfill 1% of the time).
First, we need to understand the concept of the standard normal distribution. The standard normal distribution has a mean of 0 and a standard deviation of 1. We can convert our problem into a standard normal distribution by using the z-score formula:
z = (x - μ) / σ
where z is the z-score, x is the value, μ is the mean, and σ is the standard deviation.
In our case, we want to find the z-score that corresponds to the top 1%. Since the normal distribution is symmetric, we can find the value that corresponds to the bottom 1% and then use that to find the top 1%.
To find the z-score that corresponds to the bottom 1%, we need to find the z-score that corresponds to an area of 0.01 in the left tail of the standard normal distribution. We can use a table or statistical software to find this value.
Using a standard normal distribution table or statistical software, we find that the z-score corresponding to an area of 0.01 in the left tail is approximately -2.33.
Now, we can use the z-score formula to find the value that corresponds to this z-score in our original distribution:
z = (x - μ) / σ
-2.33 = (x - μ) / 0.03
Rearranging the equation, we can solve for the mean amount of coffee, μ:
μ = x - 2.33 * 0.03
Since we want to dispense 8 ounces of coffee on average, we can substitute x with 8:
μ = 8 - 2.33 * 0.03
Calculating this gives us:
μ ≈ 7.93
Therefore, you should set the mean amount of coffee to be dispensed at approximately 7.93 ounces in order to allow the cup to overfill 1% of the time.