Question

The mean cost of a five pound bag of shrimp is 40 dollars with a standard deviation of 7 dollars. If a sample of 51 bags of shrimp is randomly selected, what is the probability that the sample mean would differ from the true mean by more than 0.6 dollars? Round your answer to four decimal places.

Answer 1

Answer:

$P(|\bar{x}-40| > 0.6)=0.5404$

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 40 dollars

Standard Deviation, σ = 7 dollars

Sample size,n = 51

We are given that the distribution of cost of shrimp is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{7}{\sqrt{51}} = 0.9801$

P(sample mean would differ by true mean by more than 0.6)

$P(|\bar{x}-40| > 0.6)\\\Rightarrow = 1-P(39.4<x<40.6)$

$P(39.4 \leq x \leq 40.6) \\\\= P(\displaystyle\frac{39.4 - 40}{0.9801} \leq z \leq \displaystyle\frac{40.6-40}{0.9801})\\\\ = P(-0.6121 \leq z \leq 0.6121)\\= P(z \leq 0.6121) - P(z < -0.6121)\\= 0.7298 - 0.2702 =0.4596$

$P(|\bar{x}|-40 > 0.6)\\= 1-0.4596=0.5404$

0.5404 is the required probability.