Suppose a brewery has a filling machine that fills 12 ounce bottles of beer. It is known that the amount of beer poured by this filling machine follows a normal distribution with a mean of 12.41 ounces and a standard deviation of 0.04 ounce. Find the probability that a randomly selected bottle contains between 12.31 and 12.37 ounces.
Last time I taught this many years ago, we still used tables or charts of "normal distribution values"
We would have to convert everything to z-scores and then look up values.
Here is a little applet that has taken the drudgery away from that
http://www.analyzemath.com/statistics/normal_calculator.html
just enter 12.41 for the mean, .04 for SD, and click on 'between' and enter the values of 12.31 and 12.37
I get .1524
I have used
http://davidmlane.com/hyperstat/z_table.html
in the past, it is really good, and shows the graph, but lately I keep getting an error when I try to enter the SD.
Perhaps it will work for you
Thank you Reiny, your comments are always helpful. I learn everytime with you.
To find the probability that a randomly selected bottle contains between 12.31 and 12.37 ounces, we need to find the area under the normal distribution curve between these two values.
First, let's calculate the z-scores for these two values using the formula:
z = (x - μ) / σ
where x is the value we want to find the z-score for, μ is the mean, and σ is the standard deviation.
For 12.31 ounces:
z1 = (12.31 - 12.41) / 0.04
z1 = -0.10 / 0.04
z1 = -2.5
For 12.37 ounces:
z2 = (12.37 - 12.41) / 0.04
z2 = -0.04 / 0.04
z2 = -1
Next, we need to find the area under the normal distribution curve between these two z-scores, which represents the probability.
Using a standard normal distribution table or a calculator, we can find the proportion of data falling below each z-score:
Area to the left of z1 = 0.0062 (from the z-score table)
Area to the left of z2 = 0.1587 (from the z-score table)
To find the area between these two z-scores, we subtract the smaller area from the larger area:
Area between z1 and z2 = Area to the left of z2 - Area to the left of z1
Area between z1 and z2 = 0.1587 - 0.0062
Area between z1 and z2 = 0.1525
Therefore, the probability that a randomly selected bottle contains between 12.31 and 12.37 ounces is approximately 0.1525, or 15.25%.
To find the probability that a randomly selected bottle contains between 12.31 and 12.37 ounces, we can use the normal distribution and z-scores.
Step 1: Calculate the z-scores for the given values.
The z-score formula is z = (X - μ) / σ , where X is the observed value, μ is the mean, and σ is the standard deviation.
For 12.31 ounces:
z1 = (12.31 - 12.41) / 0.04
For 12.37 ounces:
z2 = (12.37 - 12.41) / 0.04
Step 2: Look up the z-score probabilities in the standard normal distribution table or use a calculator with the normal distribution function.
The probability that a randomly selected bottle contains less than 12.31 ounces can be found by looking up the z1 value in the standard normal distribution table or using a calculator. Let's say this probability is P1.
The probability that a randomly selected bottle contains less than 12.37 ounces can be found by looking up the z2 value in the standard normal distribution table or using a calculator. Let's say this probability is P2.
Step 3: Calculate the final probability between the two values using the z-score probabilities.
The probability that a randomly selected bottle contains between 12.31 and 12.37 ounces is equal to P2 - P1.
Note: Make sure to use the cumulative probability (area to the left of the z-score) when looking up the z-scores in the standard normal distribution table or when using a calculator.
By following these steps, you can find the probability that a randomly selected bottle contains between 12.31 and 12.37 ounces.