Question

The distribution of actual weights of wedges of goat cheese produced at a dairy is Normal with mean 6.1 ounces and standard deviation 0.5 ounces. A SRS of 49 of these cheese wedges is selected.What is the probability that the sample mean will be greater than 6 ounces?

Answer 1

Step 1:

`\begin{gathered} \text{Given data} \\ \sigma\text{ = standard deviation = 0.5} \\ \mu\text{ = mean = 6.1} \\ \text{x = 6} \\ \text{n = 49} \end{gathered}`

Step 2:

Find the z-score

`\begin{gathered} z-score\text{ = }\frac{x\text{ - }\mu}{(\sigma)/(√(n))} \\ z-score\text{ = }\frac{√(n)(x\text{ - }\mu)}{\sigma} \\ z-score\text{ = }(√(49)(6-6.1))/(0.5) \\ z-score\text{ = }(7*-0.1)/(0.5) \\ z-score\text{ = -1.4} \end{gathered}`

Step 3:

Draw the normal distribution curve

Pr( z > -1.4) = p(z=1.4) + p(z = 0)

= 0.4192 + 0.5

= 0.9192

Final answer

What is the probability that the sample mean will be greater than 6 ounces

= .9192