agm9801

04.08.2019 •

Mathematics

Aperson travels for 1 hour and then stops for 10 minutes. the person then travels for 2 hours, then stops for 10 minutes; travels for 3 hours, stops for 10 minutes; and travels for 1 more hour. the entire trip is 357 miles. what is the person's average rate of speed while traveling? will give brainliest

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Reward18

07.01.2021

Mathematics

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"

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

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

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

P(X 2.725 + 2*5.540) = P(X13.62) = 1-P(X

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

P(X

Step-by-step explanation:

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:

$F(X) = 1- e^{-\lambda x}, x\ geq 0$

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"

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

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

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(X13.62) = 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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