Mathematics, 15.04.2020 02:41 alyssagonzales2021

Data collected at an airport suggests that an exponential distribution with mean value 2.375 hours is a good model for rainfall duration.

a. What is the probability that the duration of a particular rainfall event at this location is at least 2 hours? At most 3 hours? Between 2 and 3 hours?
b. What is the probability that rainfall duration exceeds the mean value by more than 3 standard deviations?
c. What is the probability that it is less than the mean value by more than one standard deviation?

Answers: 3

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Answer from: aliyahmuhammad5197

(a) P (X ≥ 2) = 0.4308, P (X ≤ 3) = 0.7172, P (2 ≤ X ≤ 3) = 0.1480

(b) The probability that rainfall duration exceeds the mean value by more than 3 standard deviations is 0.0183.

Step-by-step explanation:

Let X = rainfall duration (in hours).

The random variable X is exponentially distributed.

The mean duration of rainfall is:

$E (X)=2.375\ hours$

Then the parameter of the exponential distribution, λ is:

$\lambda=\frac{1}{E(X)}=\frac{1}{2.375}$

The probability distribution function of X is:

$f_{X}(x)=\lambda e^{-\lambda x};\ \lambda0,\; \; x\geq 0$

(a)

Compute the probability that the duration of a particular rainfall event at this location is at least 2 hours as follows:

$P(X\geq 2)=\int\limits^{\infty}_{2}{\frac{1}{2.375}e^{-\frac{x}{2.375}}}\, dx$

$=\frac{1}{2.375}\times \int\limits^{\infty}_{2}{e^{-\frac{x}{2.375}}}\, dx$

$=\frac{1}{2.375}\times2.375\times |e^{-\frac{x}{2.375}}|^{\infty}_{2}$

$=e^{-\frac{2}{2.375}}\\=0.4308$

Thus, the probability that the duration of a particular rainfall event at this location is at least 2 hours is 0.4308.

Compute the probability that the duration of a particular rainfall event at this location is at most 3 hours as follows:

$P(X\leq 3)=\int\limits^{3}_{0}{\frac{1}{2.375}e^{-\frac{x}{2.375}}}\, dx$

$=\frac{1}{2.375}\times \int\limits^{3}_{0}{e^{-\frac{x}{2.375}}}\, dx$

$=\frac{1}{2.375}\times2.375\times |e^{-\frac{x}{2.375}}|^{3}_{0}$

$=1-e^{-\frac{3}{2.375}}\\=1-0.2828\\=0.7172$

Thus, the probability that the duration of a particular rainfall event at this location is at most 3 hours is 0.7172.

Compute the probability that the duration of a particular rainfall event at this location is between 2 and 3 hours as follows:

P (2 ≤ X ≤ 3) = P (X ≤ 3) - P (X ≤ 2)

= P (X ≤ 3) - [1 - P (X ≥ 2)]

= 0.7172 - 1 + 0.4308

= 0.1480

Thus, the probability that the duration of a particular rainfall event at this location is between 2 and 3 hours is 0.1480.

(b)

Let a be the value that exceeds the mean value by more than 3 standard deviations.

The mean and standard deviation of an exponential distribution are same.

That is,

$a=\mu+3\sigma\\=2.375+(3\times 2.375)\\=9.5$

Compute the value of P (X > a) as follows:

P(Xa)=P(X9.5)

$=\int\limits^{\infty}_{9.5}{\frac{1}{2.375}e^{-\frac{x}{2.375}}}\, dx$

$=\frac{1}{2.375}\times \int\limits^{\infty}_{9.5}{e^{-\frac{x}{2.375}}}\, dx$

$=\frac{1}{2.375}\times2.375\times |e^{-\frac{x}{2.375}}|^{\infty}_{9.5}$

$=e^{-\frac{9.5}{2.375}}\\=0.0183$

Thus, the probability that rainfall duration exceeds the mean value by more than 3 standard deviations is 0.0183.

(c)

Let b be the value that less than the mean value by more than one standard deviation.

That is,

$b=\mu-\sigma\\=2.375-2.375\\=0$

Compute the value of P (X < b) as follows:

P(X

$=1-PX0)\\=1-\int\limits^{\infty}_{0}{\frac{1}{2.375}e^{-\frac{x}{2.375}}}\, dx$

$=1-[\frac{1}{2.375}\times \int\limits^{\infty}_{0}{e^{-\frac{x}{2.375}}}\, dx]$

$=1-[\frac{1}{2.375}\times2.375\times |e^{-\frac{x}{2.375}}|^{\infty}_{0}]$

$=1-e^{-\frac{0}{2.375}}\\=1-1\\=0$

Thus, the probability that it is less than the mean value by more than one standard deviation is 0.

Answer from: Quest

s + (a + 178) = 2000

step-by-step explanation:

2000 - 178 = 1822

1822 / 2 = 911

911 + 178 = 1089

1089 + 911 = 2000

abha = 1089

shane = 911

Answer from: Quest

answer is 144

step-by-step explanation:

$(x^{\frac{1}{8} }) (x^{\frac{3}{8} }) = 12\\$

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Data collected at an airport suggests that an exponential distribution with mean value 2.375 hours i...