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] .

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

ian gone lie this got me confused

step-by-step explanation: