Question

A random sample of checkout lines at Costca found the following information concerning length of time waiting in line (in minutes): 3.8 2.5 3.9 4.2 7.1 6.5 7.3 8.2 7.3 3.8 5.3 2.3

A. Find the mean and standard deviation for this sample.

B. Find a 99% confidence interval for the population average µ (assume  = 2):

Answer 1

Solation: A randan Sample of checka.it Oist lines at Rexired information its giren by, 3. 8. 0.5, 3.9, 4.0, 71,6.5,1.8, 8.1, 4.3.8563,9.3 @ Find the mean and Gandard deviation for this Sample.in Inter have to obtained the areas and Glandard deviation for this Sample its, Elere, Mean is xo SX [3.672.5+8.9+(1.877.1+6.5+437 8.847.6+ 3.8+5.3+2:3) 12 62.२ х 5.183 standard deviation its Sa (Etxir & V 6-1)

3 .8 - 5. 832-Eo 5 - 5.168.1.3.4- 5. Baº Cao – 5-6 - GE ++ ++-5.183324 26.5- 5.163)764-3 - 5.183)763.8 - 5.18332 .183)+64.3-5.183)+ [5.8-5.1 183776-3-5.183) tt B. o [104] 45.4367 11 (4-1306obo is - २.०324 Therefore Mlean = X= 5.103 3.0324 (standard deviation ® Find a aail. of Confidence interval for the population lake hale to obtained the act of Confidence intercal for the population average le its Here, >aat- of C.C= x +Ecritical **] ] -@ dfast S 10-1 cobere , asi-C.P =lmaq. (duf c1-0.99 as o: 01 =0.005 x19-to.oille

3 critical = trala tro.005 - 19.5758 from en I we get = 5.1868 † 8:5758%) 5.183+ [ 5.1516 rig 5.106 † [1.41671) 5.183 - 1.487 5.188 +1.487 ] ) [3.696, 6.670] Therefore, <l< 6.6 to 29.1. of C.I= (3.696