Assume 20 oz. cola bottles are filled so that the mean amount of liquid contained in the bottle is 20 oz and the standard deviation is .12 oz. If a case of 24 bottles is randomly selected from the population of all soda bottles, find the probability that the mean number of ounces in each bottle in the sample is above 20.21 (i.e., find the probability for the sample mean).
I'm a little lost. Here's what I have so far:
24 - 20.21 / .12
3.79 / .12 = 31.5833
I'm positive I've already taken a wrong turn somewhere because I'm pretty sure my z-score shouldn't be so high...
Divide by the square root of the sample size in this case since the problem is looking to find the probability for the sample mean.
20 - 20.21 / .12/√24 -->note the sample size is 24.
I hope this will help.
To find the probability that the mean number of ounces in each bottle in the sample is above 20.21, we need to standardize the sample mean using the z-score formula and then find the corresponding probability.
The formula to calculate the z-score for the sample mean is:
z = (x̄ - μ) / (σ / √n)
where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, we are given:
x̄ = 20.21 (sample mean)
μ = 20 (population mean)
σ = 0.12 (population standard deviation)
n = 24 (sample size)
Plugging in the values into the formula, we get:
z = (20.21 - 20) / (0.12 / √24)
= 0.21 / (0.12 / 4.89897949)
= 0.21 / 0.0489897949
≈ 4.2857 (rounded to 4 decimal places)
Now, to find the probability that the z-score is above 4.2857, we can consult the standard normal distribution table or use a calculator or software that provides the cumulative distribution function (CDF) for the standard normal distribution.
Using the standard normal distribution table, we can look up the probability corresponding to a z-score of 4.2857:
P(z > 4.2857) ≈ 1 - P(z ≤ 4.2857)
The standard normal distribution table typically provides probabilities for z-scores up to 3. To estimate the probability for a z-score beyond that, we can use the approximation that P(z > 3) ≈ 0.001.
Therefore, P(z > 4.2857) ≈ 1 - 0.999
≈ 0.001
So the probability that the mean number of ounces in each bottle in the sample is above 20.21 is approximately 0.001 or 0.1%.