You measure 32 turtles' weights, and find they have a mean weight of 73 ounces. Assume the population standard deviation is 7.9 ounces. Based on this, what is the maximal margin of error associated with a 95% confidence interval for the true population mean turtle weight. Give your answer as a decimal, to two places

Margin of error = 2.74 ounces  

Step-by-step explanation:

We are given the following in the question:

Sample mean, = 73 ounces

Sample size, n = 32

Alpha, α = 0.05

Population standard deviation, σ = 7.9 ounces

Margin of error:

Putting the values, we get,

Thus, the maximum margin of error associated with a 95% confidence interval for the true population mean turtle weight is 2.74 ounces

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts