Respuesta :
Answer:
The probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.
Step-by-step explanation:
Let the random variable X represent the time a child spends waiting at for the bus as a school bus stop.
The random variable X is exponentially distributed with mean 7 minutes.
Then the parameter of the distribution is,[tex]\lambda=\frac{1}{\mu}=\frac{1}{7}[/tex].
The probability density function of X is:
[tex]f_{X}(x)=\lambda\cdot e^{-\lambda x};\ x>0,\ \lambda>0[/tex]
Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:
[tex]P(6\leq X\leq 9)=\int\limits^{9}_{6} {\lambda\cdot e^{-\lambda x}} \, dx[/tex]
[tex]=\int\limits^{9}_{6} {\frac{1}{7}\cdot e^{-\frac{1}{7} \cdot x}} \, dx \\\\=\frac{1}{7}\cdot \int\limits^{9}_{6} {e^{-\frac{1}{7} \cdot x}} \, dx \\\\=[-e^{-\frac{1}{7} \cdot x}]^{9}_{6}\\\\=e^{-\frac{1}{7} \cdot 6}-e^{-\frac{1}{7} \cdot 9}\\\\=0.424373-0.276453\\\\=0.14792\\\\\approx 0.148[/tex]
Thus, the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.
Using the exponential distribution, it is found that there is a 0.1479 = 14.79% probability that the child must wait between 6 and 9 minutes on the bus on a given morning.
The exponential probability distribution, with mean m, is described by the following equation:
[tex]f(x) = \mu e^{-\mu x}[/tex]
In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.
The probability that x is lower or equal to a is given by:
[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]
Which has the following solution:
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
In this problem, we have that mean of 7 minutes, hence:
[tex]m = 7, \mu = \frac{1}{7}[/tex]
The probability that the child must wait between 6 and 9 minutes on the bus on a given morning is:
[tex]P(6 \leq X \leq 9) = P(X \leq 9) - P(X \leq 6)[/tex]
In which:
[tex]P(X \leq 9) = 1 - e^{-\frac{9}{7}} = 0.7235[/tex]
[tex]P(X \leq 6) = 1 - e^{-\frac{6}{7}} = 0.5756[/tex]
Hence:
[tex]P(6 \leq X \leq 9) = P(X \leq 9) - P(X \leq 6) = 0.7235 - 0.5756 = 0.1479[/tex]
0.1479 = 14.79% probability that the child must wait between 6 and 9 minutes on the bus on a given morning.
A similar problem is given at https://brainly.com/question/17039711
