Question

question 5 of 7
a 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:

Step1: Define z-score formula

For an individual value $X$, the z-score is $z = \frac{X - \mu}{\sigma}$

Step2: Calculate bounds for z-scores

We want $P(\mu -1 \leq X \leq \mu +1)$. Substitute $\mu=298$, $\sigma=3$:
Lower z: $z_1 = \frac{298 -1 - 298}{3} = \frac{-1}{3} \approx -0.3333$
Upper z: $z_2 = \frac{298 +1 - 298}{3} = \frac{1}{3} \approx 0.3333$

Step3: Find corresponding probabilities

We need $P(-0.3333 \leq Z \leq 0.3333) = P(Z \leq 0.3333) - P(Z \leq -0.3333)$
Using standard normal table:
$P(Z \leq 0.33) = 0.6293$, $P(Z \leq 0.34)=0.6331$; interpolating for 0.3333 gives ~0.6305
$P(Z \leq -0.3333) = 1 - P(Z \leq 0.3333) \approx 1 - 0.6305 = 0.3695$

Step4: Compute final probability

$P(-0.3333 \leq Z \leq 0.3333) = 0.6305 - 0.3695 = 0.2610$

Answer:

0.2610