The data below gives the mean price (in cents) of a liter of regular gasoline at self-service filling stations at a sample of six urban centers in Canada in May 2011. Urban area Price/liter
St. John, NL 137.1
Halifax 133.6
Montreal 136.2
Regina 132.3
Saskatoon 132.2
Yellowknife 138.9

Required:
a. Find the sample mean (in cents).
b. Find the sample standard deviation (in cents).
c. Find a 90% confidence interval for the mean gasoline price per liter across urban Canada in May 2011. Provide the lower and upper limits for your confidence interval.
d. Find a 95% confidence interval for the mean gasoline price per liter across urban Canada in May 2011. Provide the lower and upper limits for your confidence interval.

135.05 ; 2.762 ; (133.201, 136.899) ; (132.840, 137.260)

Step-by-step explanation:

Given the data:

Urban area Price/liter

St. John, NL 137.1

Halifax 133.6

Montreal 136.2

Regina 132.3

Saskatoon 132.2

Yellowknife 138.9

Mean = (137.1 + 133.6 + 136.2 + 132.3 + 132.2 + 138.9) / 6

= 810.3 / 6

= 135.05

Standard deviation (σ) : √Σ(X - mean)²/n-1

(137.1 - 135.05)^2 + (133.6 - 135.05)^2 + (136.2 - 135.05)^2 + (132.3 - 135.05)^2 + (132.2 - 135.05)^2+ (138.9 - 135.05)^2 = 38.135

√(38.135 / (6 - 1))

= √7.627

= 2.7617023

= 2.762

C.)

90% confidence interval of the mean :

Mean ± Zcrit * σ/√n

Zcrit at 90% = 1.64

135.05 ± 1.64 * (2.762/√6)

135.05 ± 1.64 * 1.1275817

Lower Value = (135.05 - (1.64 * 1.1275817)) = 133.201

Upper Value = (135.05 + (1.64 * 1.1275817))) = 136.899

Confidence interval = (133.201, 136.899)

D.)

95% confidence interval of the mean :

Mean ± Zcrit * σ/√n

Zcrit at 95% = 1.96

135.05 ± 1.96 * (2.762/6)

135.05 ± 1.96 * 1.1275817

Lower Value = (135.05 - (1.96 * 1.1275817)) = 132.840

Upper Value = (135.05 + (1.96 * 1.1275817)) = 137.260

Confidence interval = (132.840, 137.260)

answer: a video game has three difficulty levels: easy, medium, and expert. the game's scoring works in such a way that a player has to score a fixed number of points to advance to the next level, and then the scoring starts over. the total score needed to clear all three difficulty levels is 7,500. two times the number of points needed to clear the easy level is 500 more than the number of points needed to clear the medium level. twice the number of points needed to clear the medium difficulty level is 1,500 more than the number of points needed to clear the expert level. if x, y, and z represent the easy, medium, and expert levels, respectively, and a is the coefficient matrix of the system of equations modeling this situation, identify the inverse matrix, a-1, and the solution matrix, x, of the system of equations.

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