Exhibit t VariableN Mean Median TrMean StDey E Group A135 51.44 51.00 51.40 09 G

Question

Exhibit t VariableN Mean Median TrMean StDey E Group A135 51.44 51.00 51.40 09 Group B108 51.80 53.00 5205 Variable Minimum Maximum 03 Group 18.00 71.00 44.00 57.00 Group B 13.00 81.00 43.0061.00 Gven the Minitsb Output of the grades of two groups of students, answer the following: Find and. Find the probability that A scores are greater than 60. Find the probablity that scores are less than 60 Find a 91% confidence interval for Group B. Interpret. Find a 99% condence interval for Group A. Interpret. Find the probability that B scores are greater than A scares · * . Find the probability that the combined score is grester than 100

Explanation / Answer

1)
* = std.dev./sqrt(n) = 9.37/sqrt(135) = 0.8064

** = SE*sqrt(n) = 1.168*sqrt(108) = 12.1382

2)
P(X > 60) = P(z > (60 - 51.44)/0.8064) = P(z > 10.6151) = 0

3)
P(X < 60) = P(z < (60 - 51.8)/1.168) = P(z < 7.0205) = 1

4)

The above CI indicates if a sample is drawn from the population B of size 108, there are 91% chance that mean will fall within the CI.

CI for 91% n 108 mean 51.8 z-value of 91% CI 1.6954 std. dev. 12.1382 SE = std.dev./sqrt(n) 1.16800 ME = z*SE 1.98022 Lower Limit = Mean - ME 49.81978 Upper Limit = Mean + ME 53.78022 91% CI (49.8198 , 53.7802 )

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