The manager of The Cheesecake Factory in Boston reports that on six randomly selected weekdays, the number of customers served was 175, 125, 180, 220, 240, and 245. She believes that the number of customers served on weekdays follows a normal distribution. Construct the 99% confidence interval for the average number of customers served on weekdays.

(121.576 ; 273.424)

Step-by-step explanation:

Given the data:

175, 125, 180, 220, 240, 245

We can calculate the mean and standard deviation

Mean = Σx/ n = 1185 / 6 = 197.5

Standard deviation = 46.125 (calculator)

The confidence interval :

Mean ± margin of error

Margin of Error = Tcritical * s/sqrt(n)

Tcritical at 99%, df = n - 1 ; 6 - 1 = 5

Tcritical = 4.032

Margin of Error = 4.032 * 46.125/√6

Margin of error = 75.924

Confidence interval :

197.5 ± 75.924

Lower boundary = 197.5 - 75.924 = 121.576

Upper boundary = 197.5 + 75.924 = 273.424

(121.576 ; 273.424)

yes because they have the same amount of green and blur marbles.

step-by-step explanation: