Solution
Step 1:
[tex]\begin{gathered} \text{Given data} \\ \sigma\text{ = standard deviation = 0.5} \\ \mu\text{ = mean = 6.1} \\ \text{x = 6} \\ \text{n = 49} \end{gathered}[/tex]Step 2:
Find the z-score
[tex]\begin{gathered} z-score\text{ = }\frac{x\text{ - }\mu}{\frac{\sigma}{\sqrt{n}}} \\ z-score\text{ = }\frac{\sqrt{n}(x\text{ - }\mu)}{\sigma} \\ z-score\text{ = }\frac{\sqrt{49}(6-6.1)}{0.5} \\ z-score\text{ = }\frac{7\times-0.1}{0.5} \\ z-score\text{ = -1.4} \end{gathered}[/tex]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
