Respuesta :
Answer:
(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.
(c) The probability that it is less than the mean value by more than one standard deviation is 0.
Step-by-step explanation:
Let X = rainfall duration (in hours).
The random variable X is exponentially distributed.
The mean duration of rainfall is:
[tex]E (X)=2.375\ hours[/tex]
Then the parameter of the exponential distribution, λ is:
[tex]\lambda=\frac{1}{E(X)}=\frac{1}{2.375}[/tex]
The probability distribution function of X is:
[tex]f_{X}(x)=\lambda e^{-\lambda x};\ \lambda>0,\; \; x\geq 0[/tex]
(a)
Compute the probability that the duration of a particular rainfall event at this location is at least 2 hours as follows:
[tex]P(X\geq 2)=\int\limits^{\infty}_{2}{\frac{1}{2.375}e^{-\frac{x}{2.375}}}\, dx[/tex]
[tex]=\frac{1}{2.375}\times \int\limits^{\infty}_{2}{e^{-\frac{x}{2.375}}}\, dx[/tex]
[tex]=\frac{1}{2.375}\times2.375\times |e^{-\frac{x}{2.375}}|^{\infty}_{2}[/tex]
[tex]=e^{-\frac{2}{2.375}}\\=0.4308[/tex]
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:
[tex]P(X\leq 3)=\int\limits^{3}_{0}{\frac{1}{2.375}e^{-\frac{x}{2.375}}}\, dx[/tex]
[tex]=\frac{1}{2.375}\times \int\limits^{3}_{0}{e^{-\frac{x}{2.375}}}\, dx[/tex]
[tex]=\frac{1}{2.375}\times2.375\times |e^{-\frac{x}{2.375}}|^{3}_{0}[/tex]
[tex]=1-e^{-\frac{3}{2.375}}\\=1-0.2828\\=0.7172[/tex]
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,
[tex]a=\mu+3\sigma\\=2.375+(3\times 2.375)\\=9.5[/tex]
Compute the value of P (X > a) as follows:
[tex]P(X>a)=P(X>9.5)[/tex]
[tex]=\int\limits^{\infty}_{9.5}{\frac{1}{2.375}e^{-\frac{x}{2.375}}}\, dx[/tex]
[tex]=\frac{1}{2.375}\times \int\limits^{\infty}_{9.5}{e^{-\frac{x}{2.375}}}\, dx[/tex]
[tex]=\frac{1}{2.375}\times2.375\times |e^{-\frac{x}{2.375}}|^{\infty}_{9.5}[/tex]
[tex]=e^{-\frac{9.5}{2.375}}\\=0.0183[/tex]
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,
[tex]b=\mu-\sigma\\=2.375-2.375\\=0[/tex]
Compute the value of P (X < b) as follows:
[tex]P(X<b)=P(X<0)[/tex]
[tex]=1-PX>0)\\=1-\int\limits^{\infty}_{0}{\frac{1}{2.375}e^{-\frac{x}{2.375}}}\, dx[/tex]
[tex]=1-[\frac{1}{2.375}\times \int\limits^{\infty}_{0}{e^{-\frac{x}{2.375}}}\, dx][/tex]
[tex]=1-[\frac{1}{2.375}\times2.375\times |e^{-\frac{x}{2.375}}|^{\infty}_{0}][/tex]
[tex]=1-e^{-\frac{0}{2.375}}\\=1-1\\=0[/tex]
Thus, the probability that it is less than the mean value by more than one standard deviation is 0.
