Respuesta :
Answer:
The probability that a car driver can complete the transaction in less than 5 minutes is 0.4647.
Step-by-step explanation:
Let X = times required by drivers to fill their car's tank and pay.
The average time required by drivers is, β = 8 minutes.
The random variable X is Exponentially distributed with parameter, [tex]\lambda=\frac{1}{\beta}=\frac{1}{8}=0.125[/tex].
The probability distribution function of exponential distribution is:
[tex]f_{X}(x)=\lambda e^{-\lambda x};\ x>0[/tex]
Compute the probability that a car driver can complete the transaction in less than 5 minutes as follows:
[tex]P(X<5)=\int\limits^{5}_{0} {\lambda e^{-\lambda x}} \, dx\\=\lambda \int\limits^{5}_{0} {e^{-\lambda x}} \, dx\\=\lambda |\frac{e^{-\lambda x}}{-lambda}|^{5}_{0} \\=|-e^{-\lambda x}|^{5}_{0}\\=-e^{-0.125\times 5}+e^{-0.125\times 0}\\=1-0.5353\\=0.4647[/tex]
Thus, the probability that a car driver can complete the transaction in less than 5 minutes is 0.4647.
Answer:
Probability that a car can complete the transaction in less than 5 minutes is 0.465.
Step-by-step explanation:
We are given that the manager of a gas station has observed that the times required by drivers to fill their car's tank and pay are quite variable. In fact, the times are exponentially distributed with a mean of 8 minutes.
Let X = times required by drivers to fill their car's tank and pay
The probability distribution function of exponential distribution is given by;
[tex]f(x) = \lambda e^{-\lambda x} , x >0[/tex] Â Â where, [tex]\lambda[/tex] = parameter of distribution.
Now, the mean of exponential distribution is = [tex]\frac{1}{\lambda}[/tex]  which is given to us as 8 minutes that means  [tex]\lambda = \frac{1}{8}[/tex]  .
So, X ~ Exp( [tex]\lambda = \frac{1}{8}[/tex] )
Also, we know that Cumulative distribution function (CDF) of Exponential distribution is given as;
[tex]F(x) = P(X \leq x) = 1 - e^{-\lambda x}[/tex] Â , x > 0
Now, Probability that a car can complete the transaction in less than 5 minutes is given by = P(X < 5 min)
 P(X < 5 min) = [tex]1 - e^{-\frac{1}{8} \times 5}[/tex]     {Using CDF}
            = 1 - 0.5353 = 0.465
Therefore, probability that a car can complete the transaction in less than 5 minutes is 0.465.
