Answer:
The answer is "".
Step-by-step explanation:
Please find the complete question in the attached file.
We select a sample size n from the confidence interval with the mean [tex]\mu[/tex]and default [tex]\sigma[/tex], then the mean take seriously given as the straight line with a z score given by the confidence interval
[tex]\mu=3.87\\\\\sigma=2.01\\\\n=110\\\\[/tex]
Using formula:
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
The probability that perhaps the mean shells length of the sample is over 4.03 pounds is
[tex]P(X>4.03)=P(z>\frac{4.03-3.87}{\frac{2.01}{\sqrt{110}}})=P(z>0.8349)[/tex]
Now, we utilize z to get the likelihood, and we use the Excel function for a more exact distribution
[tex]=\textup{NORM.S.DIST(0.8349,TRUE)}\\\\P(z<0.8349)=0.7981[/tex]
the required probability: [tex]P(z>0.8349)=1-P(z<0.8349)=1-0.7981=\boldsymbol{0.2019}[/tex]
