Assume the distribution of commute times to a major city follows the normal probability distribution, and the standard deviation is 4.2 minutes. A random sample of 13 commute times is given below in minutes. Find the 98% confidence interval for the mean travel time in minutes. Round your answers to two decimal places and use ascending order.

The 98% CI for the mean travel time to the major city is [20.70; 26.14]minutes

Step-by-step explanation:

Hello!

The variable of interest is

X: commute time to a major city.

This variable has a normal distribution

X~N(μ;σ²)

The standard deviation is known to be:

σ= 4.2 minutes

A random sample of n= 13 commute times was taken:

11.5, 13.2, 14.7, 17.1, 18.7, 21.8, 22.4, 25, 26.9, 27.6, 31.1, 35.9, 38.6

You need to estimate the population mean travel time by calculating a 98% CI.

Since the variable has a normal distribution and the population standard deviation is known ,the statistic to use for this interval is the standard normal, then the formula for the interval is:

[X[bar] ± * ]

The value of the statistic for the 1 - α: 0.98 interval is:

Next is to calculate the sample mean:

X[bar]= ∑X/n= 304.5/13= 23.42 minutes

[23.42 ± 2.334 * ()]

[20.70; 26.14]minutes

With a 98% confidence level you'd expect that the interval [20.70; 26.14]minutes will contain the true mean travel time to the major city.

I hope this helps!

a.) list the factors of 40.

1/2/4/5/8/10/20/40

b.) list the factors of 40 that represent the number of cans that could be on each shelf.

5

c.) how many different arrangements of cans are possible?

just one because if you put more on one shelf you will go way over 40 and the cabinet will probably break.

d.) what are the possible arrangements of the cans?

put 5 cans on each self to reach 40 cans in total.

e.) which arrangement uses all the shelves?

5 because you can put 5 cans on each shelf, and since there are 8 shelfs you are equally going to get 40 cans.