Question

many breakfast bars were sampled and found to have a mean weight of 9 oz and a standard deviation of.13 oz. what percent of the breakfast bars weigh more than 9.14 oz?

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 9.14\) (the value we are interested in), \(\mu=9\) (the mean), and \(\sigma = 0.13\) (the standard deviation).

$$z=\frac{9.14 - 9}{0.13}=\frac{0.14}{0.13}\approx1.08$$

Step2: Find the probability using the standard normal distribution

We want to find \(P(X>9.14)\), which is equivalent to \(P(Z > 1.08)\) since \(X\) (weight of breakfast bars) is normally distributed.
Using the property \(P(Z>z)=1 - P(Z\leq z)\). From the standard - normal table, \(P(Z\leq1.08)=0.8599\)

$$P(Z > 1.08)=1-0.8599 = 0.1401$$

Answer:

Approximately \(14.01\%\) of the breakfast bars weigh more than \(9.14\) oz.