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? a 1% b 8% c 12% d 16% e 99%

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 35$ mg (the value we are interested in), $\mu=33$ mg (the mean), and $\sigma = 1.2$ mg (the standard deviation). So, $z=\frac{35 - 33}{1.2}=\frac{2}{1.2}\approx1.67$.

Step2: Use the standard normal distribution 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 $P(Z>z)=1 - P(Z\leq z)$. Looking up $P(Z\leq1.67)$ in the standard - normal table, we find that $P(Z\leq1.67)\approx0.9525$. Then $P(Z > 1.67)=1 - 0.9525 = 0.0475\approx5\%$. The closest option to 5% is 8% (option B).

Answer:

B. 8%