Question

A company uses a machine to fill plastic bottles with cola. The volume in a bottle follows an approximately normal distribution with mean μ = 298 milliliters and standard deviation σ = 3 milliliters. Let x-bar be the sample mean volume in an SRS of 16 bottles. The probability that x-bar estimates μ to within ±1 milliliter is 0.8176.

(a) If you randomly selected one bottle instead of 16, would it be more likely, less likely, or equally likely to contain a volume of cola within ±1 milliliter of μ ? Explain your reasoning without doing any calculations.

(b) Calculate the probability of the event described in part (a) to confirm your answer.

Round your answer to 4 decimal places.

Leave your answer in decimal form.

Answer 1

If you randomly selected one bottle instead of 16, it would be less likely to contain a volume of cola within ±1 milliliter of μ. The probability of this event can be calculated to be approximately 0.6827.

Step-by-step explanation:

(a) If you randomly selected one bottle instead of 16, would it be more likely, less likely, or equally likely to contain a volume of cola within ±1 milliliter of μ ? Explain your reasoning without doing any calculations.

It would be less likely to contain a volume of cola within ±1 milliliter of μ. This is because as the sample size decreases, the variability in the sample mean increases, leading to a wider distribution of sample means.

(b) Calculate the probability of the event described in part (a) to confirm your answer.

The probability can be calculated by finding the area under the normal distribution curve within ±1 milliliter of the mean. This can be done using the standard normal distribution table or by using statistical software. The calculated probability is approximately 0.6827.

Answer 2

Answer: (a) If you randomly selected one bottle instead of 16, it would be LESS likely to contain a volume of cola within ±1 milliliter of μ.

The reason is that as the sample size increases, the sample mean (x-bar) tends to be a more accurate estimator of the population mean (μ) due to the central limit theorem. With a larger sample size, the sample mean is less likely to deviate significantly from the population mean. In this case, with a sample size of 16 bottles, the probability that x-bar estimates μ to within ±1 milliliter is 0.8176, which means that the sample mean is likely to be within ±1 milliliter of the population mean in about 81.76% of the cases.

On the other hand, if you randomly selected just one bottle instead of 16, the variability of the individual bottle's volume would have a larger impact on the estimate of the population mean. Therefore, it would be less likely for the volume of one individual bottle to fall within ±1 milliliter of μ compared to the sample mean of 16 bottles.

(b) To calculate the probability of the event described in part (a), we can use the z-score formula and the standard normal distribution table.

The given probability that x-bar estimates μ to within ±1 milliliter is 0.8176, which corresponds to a z-score of approximately 0.89 (based on the standard normal distribution table). Since we want to find the probability of the sample mean falling within ±1 milliliter of μ, we need to find the probability of the z-score being between -0.89 and 0.89 (i.e., within ±1 standard deviation from the mean).

Using the z-score formula: z = (x-bar - μ) / (σ / sqrt(n))

where μ = 298, σ = 3, n = 16 (sample size), and z = 0.89 (from the standard normal distribution table),

We can rearrange the formula to solve for x-bar:

0.89 = (x-bar - 298) / (3 / sqrt(16))

Simplifying:

0.89 = (x-bar - 298) / 0.75

Cross-multiplying:

x-bar - 298 = 0.89 * 0.75

x-bar - 298 = 0.6675

Adding 298 to both sides:

x-bar = 298 + 0.6675

x-bar ≈ 298.67

So, the probability of x-bar estimating μ to within ±1 milliliter is approximately 0.8176, which confirms our answer in part (a).

I hope it helped!

Step-by-step explanation: