Question

the caffeine content of 8 - ounce cans of a certain cola drink is approximately normally distributed with mean 33 milligrams (mg). a randomly selected 8 - ounce can containing 35 mg of caffeine is 1.2 standard deviations above the mean. approximately what percent of 8 - ounce cans of the cola have a caffeine content greater than 35 mg?

Explanation:

Step1: Calculate z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 35$, $\mu=33$, and $\sigma = 1.2$. So $z=\frac{35 - 33}{1.2}=\frac{2}{1.2}\approx1.67$.

Step2: Use z - table

We want to find $P(X>35)$, which is equivalent to $P(Z > 1.67)$ in the standard normal distribution. Since the total area under the standard - normal curve is 1, and $P(Z\leq z)$ can be found from the z - table. Looking up $z = 1.67$ in the z - table, we get $P(Z\leq1.67)=0.9525$. Then $P(Z > 1.67)=1 - P(Z\leq1.67)=1 - 0.9525 = 0.0475\approx5\%$. The closest option is B. 8%.

Answer:

B. 8%