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 52 ounces and the standard deviation is 8.4 ounces. What is the lower bound of the 99 percent confidence interval for the gift basket's average shipping weight

The lower bound of the 99% confidence interval for the gift basket's average shipping weight is of 49.34 ounces.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 70 - 1 = 69

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 69 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.649

The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 52 - 2.66 = 49.34

The lower bound of the 99% confidence interval for the gift basket's average shipping weight is of 49.34 ounces.

example 2 – simplify: step 1: completely factor both the numerators and denominators of all fractions. step 2: change the division sign to a multiplication sign and flip (or reciprocate) the fraction after the division sign; essential you need to multiply by the reciprocal. step 3 : cancel or reduce the fractions.

total amount to be paid = $19.5

step-by-step explanation:

given ,

original price = $25.99

discount = 25%

discount amount = 25% of original price

discount amount = 25/100   * 25.99

discount amount = $6.49

hence , the total amount to be paid = original price - discount amount

                                                            = $25.99 - $6.49

                                                            =$19.5

total amount to be paid = $19.5