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Novosadov [1.4K]
1 year ago
6

What is the probability that the commute on a particular game day exceeds the commute on a particular non-game day?

Social Studies
1 answer:
castortr0y [4]1 year ago
3 0

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

Learn more about probability here brainly.com/question/24756209

#SPJ4

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