A bottling machine can be regulated so that it discharges an average of u ounces per bottle. It has been observed that the amount of fill dispensed by the machine is Normally distributed with sigma =

Using the central limit theorem, we know that the distribution of sample means will be Normal with a mean of u and a standard deviation of {eq}frac{sigma}{sqrt{n}}=frac{0.2}{sqrt{9}}=0.0667 {/eq} ounces. Therefore, we need to find the probability that the sample mean differs from u by at most 0.3 ounce, or within the range {eq}u pm 0.3 {/eq}. We can convert this range to a z-score range by standardizing with the formula {eq}z=frac{bar{x}-u}{frac{sigma}{sqrt{n}}} {/eq}. Plugging in the values, we get {eq}z=frac{0.3}{0.0667}=4.5 {/eq} for the upper bound and {eq}z=-4.5 {/eq} for the lower bound. Using a standard Normal distribution table or a calculator, we can find the probability of being within this z-score range to be \approx imately 0.9985. Therefore, the probability that the sample mean differs from u by at most 0.3 ounce is 0.9985. When {eq}sigma {/eq} is unknown and we use the sample variance {eq}s^2=5.3 {/eq}, we need to use a t-distribution instead of a Normal distribution. The formula for the t-score is similar to the z-score, but we use the sample standard deviation {eq}s {/eq} instead of {eq}sigma {/eq}. The formula is {eq}t=frac{bar{x}-u}{frac{s}{sqrt{n}}} {/eq}. We don't know the true value of {eq}sigma {/eq}, but we can estimate it with the sample standard deviation {eq}s {/eq}. The t-distribution takes into account the added uncertainty of using an estimated value instead of the true value. Using the formula for the t-score and the sample values, we get {eq}t=frac{0.3}{frac{sqrt{5.3}}{sqrt{9}}}=2.546 {/eq} for the upper bound and {eq}t=-2.546 {/eq} for the lower bound. Using a t-distribution table or a calculator with 8 degrees of freedom (n-1), we can find the probability of being within this t-score range to be \approx imately 0.0246. Therefore, the probability that the sample mean differs from u by at most 0.3 ounce is 0.0246 when {eq}sigma {/eq} is unknown and we use the sample variance {eq}s^2=5.3 {/eq}.