Question

1. sometimes machines used to fill soda cans can overfill or underfill. for one such machine, the distribution of the amount of soda filled into 12 - ounce cans is approximately normal with a mean of 12.05 ounces and a standard deviation of 0.02 ounces. let ( s ) = the amount of soda (in ounces) in a randomly selected 12 - ounce can filled by this machine.

a. sketch the probability distribution of ( s ). be sure to label and scale the horizontal axis.
b. what is the probability of selecting a can that has been overfilled (more than 12 ounces)?
c. find ( p(12 leq s leq 12.10) ).
d. if 500 cans are randomly selected, how many of them would we expect to be underfilled? show your work.

Explanation:

Step1: Define normal distribution

$S \sim N(\mu=12.05, \sigma=0.02)$

Step2: (Part a) Label horizontal axis

Mark $\mu=12.05$ at the peak. Mark $\mu\pm\sigma$: $12.03, 12.07$; $\mu\pm2\sigma$: $12.01, 12.09$; $\mu\pm3\sigma$: $11.99, 12.11$ on the horizontal axis of the bell curve.

Step3: (Part b) Calculate z-score for 12

$z = \frac{x-\mu}{\sigma} = \frac{12-12.05}{0.02} = -2.5$

Step4: (Part b) Find right-tail probability

$P(S>12) = P(Z>-2.5) = 1 - P(Z\leq-2.5)$
From z-table, $P(Z\leq-2.5)=0.0062$, so $1-0.0062=0.9938$

Step5: (Part c) Calculate z-scores for 12 and 12.10

$z_1 = \frac{12-12.05}{0.02}=-2.5$, $z_2 = \frac{12.10-12.05}{0.02}=2.5$

Step6: (Part c) Find interval probability

$P(12\leq S\leq12.10)=P(-2.5\leq Z\leq2.5)=P(Z\leq2.5)-P(Z\leq-2.5)$
From z-table, $P(Z\leq2.5)=0.9938$, $P(Z\leq-2.5)=0.0062$, so $0.9938-0.0062=0.9876$

Step7: (Part d) Find underfill probability

Underfill is $S<12$, so $P(S<12)=P(Z<-2.5)=0.0062$

Step8: (Part d) Calculate expected number

$\text{Expected number} = 500 \times 0.0062 = 3.1$

Answer:

a. Bell curve with horizontal axis labeled: 11.99, 12.01, 12.03, 12.05 (peak), 12.07, 12.09, 12.11
b. 0.9938
c. 0.9876
d. 3 (rounded) or 3.1