Question

question 5 of 7a company uses a machine to fill plastic bottles with cola. the volume in a bottle follows an approximately normal distribution with mean $mu = 298$ milliliters and standard deviation $sigma = 3$ milliliters. let $\bar{x}$ be the sample mean volume in an srs of 16 bottles. the probability that $\bar{x}$ estimates $mu$ to within $pm 1$ 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 $pm 1$ milliliter of $mu$? explain your reasoning without doing any calculations.(b) calculate the probability of the event described in part (a) to confirm your answer.(round to 4 decimal places. leave your answer in decimal form.)

Explanation:

Part (a) (Answer-Explanation Format)

Sample means have less variability than individual data points because the sampling distribution of the mean has a smaller standard deviation (spread) than the original population distribution. This makes individual values less likely to be close to the population mean compared to sample averages.

Step1: Define the random variable

Let $X$ = volume of one bottle, where $X \sim N(\mu=298, \sigma=3)$

Step2: Set up the probability range

We need $P(298-1 \leq X \leq 298+1) = P(297 \leq X \leq 299)$

Step3: Calculate z-scores

Z-score formula: $z = \frac{X-\mu}{\sigma}$
For $X=297$: $z_1 = \frac{297-298}{3} = -\frac{1}{3} \approx -0.3333$
For $X=299$: $z_2 = \frac{299-298}{3} = \frac{1}{3} \approx 0.3333$

Step4: Find the probability

Use standard normal table: $P(-0.3333 \leq Z \leq 0.3333) = P(Z \leq 0.3333) - P(Z \leq -0.3333)$
From standard normal distribution: $P(Z \leq 0.3333) \approx 0.6305$, $P(Z \leq -0.3333) \approx 0.3695$
Subtract: $0.6305 - 0.3695 = 0.2610$

Answer:

Less likely. Individual values vary more than averages, so getting an individual value that is close to the true mean is less likely.

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Part (b) (Step-by-Step Format)