ExponentialDist
2;5
= .33 = P(X≤2), so the answer
to the second question is ``about 33 percent.''BITMAPSETProbSolvHint0.1807in0.2188in0inq1
NormalDist
2;5, 2
= .067 = P(X≤2), so the answer to
the second question is ``about 7 percent.''
![]() ![]() ![]() ![]() |
= | μ | |
![]() ![]() ![]() ![]() ![]() ![]() |
= | σ2 |
For the Student's t distribution, with five degrees of freedom, Evaluate gives
Pr![]() ![]() |
= | Pr(x≤2) - Pr(x≤ - 2) | |
= | ![]() ![]() |
||
= | .84 |