Given:
Mean, μ = 119.54 oz
Standard deviation, σ = 0.62 oz
Sample size, n = 14
Part 2.
Let's find the probability that the mean weight is at most 119.18 oz.
Here, we are to find:
P(X ≤ 119.18)
Apply the formula:
[tex]P(X\leq\bar{x})=P(z\leq\frac{\bar{x}-\mu}{E})[/tex]Where:
x' = 119.18
μ = 119.54
E = 0.1657
Thus, we have:
[tex]\begin{gathered} P(X\leq119.18)=P(z\leq\frac{119.18-119.54}{0.1657}) \\ \\ P(X\leq119.18)=P(z\leq\frac{-0.36}{0.1657}) \\ \\ P(X\leq119.18)=P(z\leq−2.1726) \end{gathered}[/tex]Using the Standard Normal Distribution Table, we have:
NORMSDIST(−2.1726) = 0.0149
Therefore, the probability that the mean weight is at most 119.18 is 0.0149
• ANSWER:
0.0149