Question

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

Answer 1

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:

`z=(x-\mu)/(\sigma)`

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.

`s=(\sigma)/(√(n))`
`\begin{gathered} \sigma=√(variance)=√(36)=6 \\ s=(6)/(√(43)) \end{gathered}`

What is the probability that the sample mean would differ from the true mean by greater than 1.4 dollar?

x = 48.4

`\begin{gathered} z=(48.4-47)/((6)/(√(43))) \\ z=1.53006 \end{gathered}`

x=45.6

`z=(45.6-47)/((6)/(√(43)))=-1.53006`
`\begin{gathered} When\text{ }z=1.53006 \\ p-value=0.9369990423 \\ when\text{ z=-1.53006} \\ p-value=0.0630009577 \end{gathered}`

The difference will be;

`0.9369990423-0.0630009577=0.8739980846`
`\begin{gathered} p+0.8739980846=1 \\ p=1-0.8739980846 \\ p=0.1260019154 \\ p\approx0.1260 \end{gathered}`

Answer;

`p=0.1260`