The required probability is 0.99.
Probability is the branch of arithmetic concerning numerical descriptions of how probably an event is to occur, or how possibly it's far that a proposition is true. The probability of an occasion is a number of between zero and 1, in which, kind of speak, 0 shows the impossibility of the event and 1 shows actuality.
Probability is the branch of arithmetic regarding numerical descriptions of ways probable an event is to arise, or how possibly it's miles that a proposition is true. The opportunity of an occasion is quite a number between 0 and 1, in which, more or less speaking, zero indicates the impossibility of the occasion and 1 shows reality.
An opportunity is a number that displays the hazard or probability that a specific occasion will arise. probabilities may be expressed as proportions that range from zero to at least one, and they also can be expressed as chances ranging from 0% to 100%.
The probability that the commutes on a particular game day exceed the commute
P(XY) = P(X-Y>0)
=1-P(X-Y≤0)
X - Y is normally distributed with mean μ=E[X-Y]=E[X]− E[Y]and
variance ²=V[X-Y] =σ₂+02
Then X-Y-N(u.o²) (8.13)
P(X-Y>0)=1-P(X-Y≤ 0)
=1-P
(X-Y)-0-8
√13
=1-P z≤
-8
3.6055
=1-P(z≤-2.22)
=1-(-2.22)
=1-(0.0132)
0.99
Therefore, the required probability is 0.99
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