Announcements for 84 upcoming engineering conferences were randomly picked from a stack of ieee spectrum magazines. the mean length of the conferences was 3.94 days, with a standard deviation of 1.28 days. assume the underlying population is normal. what is the error bound of 95% confidence interval for the population mean length of engineering conferences? construct a 95% confidence interval for the population mean length of engineering conferences. what is the lower bound? construct a 95% confidence interval for the population mean length of engineering conferences. what is the upper bound?

Step-by-step explanation:

Hello!

You need to construct a 95% CI for the population mean of the length of engineering conferences.

The variable has a normal distribution.

The information given is:

n= 84

x[bar]= 3.94

δ= 1.28

The formula for the Confidence interval is:

x[bar]±*(δ/n)

Lower bound(Lb): 3.698

Upper bound(Ub): 4.182

Error bound: (Ub - Lb)/2 = (4.182-3.698)/2 = 0.242

I hope it helps!

its a straight line parallel to the time axis

the final amount you will pay is $68.04.

the tip is determined by the pre-tax amount, so we calculate the tip based on the original bill, not after we've added sales tax.

to find the tip:

we can divide 54.00 by 100 to get 1% of 54. then we multiply that by 18.

54 / 100 = .54

.54 x 18 = 9.72

tip = 9.72$

to find the tax:

we can divide 54.00 by 100 to get 1% of 54. then we can multiply that by 8 to get 8%.

54 / 100 = .54

.54 x 8 = 4.32

tax= 4.32

now we can add the tip and tax to the bill to get our total.

54 + 9.72 + 4.32 = 68.04

total: $68.04

step-by-step explanation: