Question

cans of a cola beverage claim to contain 16 ounces. all cans are filled so that \\( \mu x = 16 \\) ounces (as labeled) and \\( \sigma x = 0.143 \\). samples of size \\( n = 34 \\) are randomly selected and the amount of cola is measured.
part a:
find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces.
(round your answer to four decimal places.)

Explanation:

Step1: Calculate the standard error

The standard error \(SE=\frac{\sigma_x}{\sqrt{n}}\). Given \(\sigma_x = 0.143\) and \(n = 34\), then \(SE=\frac{0.143}{\sqrt{34}}\approx\frac{0.143}{5.831}\approx0.0245\).

Step2: Calculate the z - score

The z - score formula is \(z=\frac{\bar{x}-\mu_x}{SE}\). Here, \(\bar{x}=16.01\), \(\mu_x = 16\), and \(SE\approx0.0245\). So \(z=\frac{16.01 - 16}{0.0245}=\frac{0.01}{0.0245}\approx0.41\).

Step3: Find the probability

We want \(P(\bar{X}>16.01)\), which is equivalent to \(P(Z > 0.41)\). Since \(P(Z>z)=1 - P(Z\leq z)\), and from the standard normal table \(P(Z\leq0.41)=0.6591\). Then \(P(Z > 0.41)=1- 0.6591=0.3409\).

Answer:

\(0.3409\)