Respuesta :
Step 1
Given; The main cost of a 5 pound bag of shrimp is $47 with a variance of 36. If a sample of 43 bags of shrimp is randomly selected what is the probability that the sample mean would differ from the true mean by greater than $1.4? round your answer to four decimal places.
Step 2
In a set with mean and standard deviation, the z-score of a measure X is given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
The Central Limit Theorem establishes that, for a random variable X, with mean and standard deviation, the sample means with size n of at least 30 can be approximated to a normal distribution with mean and standard deviation.
[tex]s=\frac{\sigma}{\sqrt{n}}[/tex][tex]\begin{gathered} \sigma=\sqrt{variance}=\sqrt{36}=6 \\ s=\frac{6}{\sqrt{43}} \end{gathered}[/tex]What is the probability that the sample mean would differ from the true mean by greater than 1.4 dollar?
x = 48.4
[tex]\begin{gathered} z=\frac{48.4-47}{\frac{6}{\sqrt{43}}} \\ z=1.53006 \end{gathered}[/tex]x=45.6
[tex]z=\frac{45.6-47}{\frac{6}{\sqrt{43}}}=-1.53006[/tex][tex]\begin{gathered} When\text{ }z=1.53006 \\ p-value=0.9369990423 \\ when\text{ z=-1.53006} \\ p-value=0.0630009577 \end{gathered}[/tex]The difference will be;
[tex]0.9369990423-0.0630009577=0.8739980846[/tex][tex]\begin{gathered} p+0.8739980846=1 \\ p=1-0.8739980846 \\ p=0.1260019154 \\ p\approx0.1260 \end{gathered}[/tex]Answer;
[tex]p=0.1260[/tex]