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Arisa [49]
3 years ago
14

The mean number of points scored by the los angeles lakers was 101.7. in what proportion of their game did the celtics score mor

e than the lakers' mean score?
Mathematics
1 answer:
Alex787 [66]3 years ago
4 0
Thank you for posting your question here at brainly. I hope the answer will help. Belowis the solution:

Let X be a random variable for the Celtics score. Then you want 

<span>P(X > 101.7) </span>
<span>  1 - P(X < 101.7) </span>
<span>  = 1 - P((X - μ)/σ < (101.7 - 99.2)/10.5)) </span>
<span>  = 1 - N_Z(0.2381) </span>
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Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

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P(X

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Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

The cumulative distribution for this function is given by:

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We know the value for the mean on this case we have that :

mean = \frac{1}{\lambda}

\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

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Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: Var(X) =\frac{1}{\lambda^2}

And the deviation would be:

Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725

And the mean is given by Mean = 2.725

Two deviations correspond to 5.540, so we want this probability:

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

P(X

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