Question 1128429
since the sample size(40) is greater than 30, we can use the Normal Distribution
:
z-score = (9.1 - 8.4) / 1.8 = 0.3889 is approximately 0.39
:
Probalbility(P) (X > 9.1) = 1 -P(X < 9.1)
:
P(X < 9.1) is the associated probability for z-score 0.39 in the table of z-scores
:
P(X < 9.1) = 0.6517
:
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P(X > 9.1) = 1 - 0.6517 = 0.3483
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