Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of 6 minutes. A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes. What parameters should be used to calculate a confidence interval at the 95% confidence level for this population mean?

95% confidence interval = [ 33.78 , 38.22 ]

Step-by-step explanation:

We are given that average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of 6 minutes.

A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes.

The Pivotal quantity for 95% confidence interval is given by;

              ~ N(0,1)

where, X bar = sample mean = 36

              = population standard deviation = 6

             n = sample size = 28

So, 95% confidence interval for population mean, is given by;

P(-1.96 < N(0,1) < 1.96) = 0.95

P(-1.96 < < 1.96) = 0.95

P(-1.96 * < < 1.96 * ) = 0.95

P(X bar - 1.96 * < < X bar + 1.96 * ) = 0.95

95% confidence interval for = [ X bar - 1.96 * , X bar + 1.96 * ]

                                               = [ 36 - 1.96 * , 36 + 1.96 * ]

                                               = [ 33.78 , 38.22 ]

Therefore, 95% confidence interval for population mean is [33.78 , 38.22] .

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet