Suppose the shipping weight of your cheese shop's customized gift baskets is asymmetrically distributed with unknown mean and standard deviation. For a sample of 70 orders, the mean weight is 57 ounces and the standard deviation is 7.1 ounces. What is the lower bound of the 90 percent confidence interval for the gift basket's average shipping weight

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The lower bound of the 90% confidence interval for the gift basket's average shipping weight is 55.605.

Given mean weight of 57 ounces ,sample size 70 ,confidence level 90% and standard deviation of 7.1 ounces.

We have to find the lower bound of the confidence interval for the gift's basket's avrage shipping weight.

We can easily find the confidence interval and its lower bound through theformula of margin of error.

Margin of error is the difference between real values and calculated values.

Margin of error=z*σ/[tex]\sqrt{n}[/tex]

where z is the critical value of confidence level

σ is standard deviation,

n is the sample size

We have to first find the z value for 90% confidence level which is 1.645.

Margin of error=1.645*7.1/[tex]\sqrt{70}[/tex]

=11.6795/8.3666

=1.395

Lower bound of the confidence interval = Mean - margin of error

=57-1.395

=55.605.

Hence the lower bound of the confidence interval for the gift basket's average shipping weight is 55.605.

Learn more about margin of error at https://brainly.com/question/10218601

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