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 <em>X</em> represent the time a child spends waiting at for the bus as a school bus stop.
The random variable <em>X</em> is exponentially distributed with mean 7 minutes.
Then the parameter of the distribution is,
.
The probability density function of <em>X</em> is:

Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:

![=\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](https://tex.z-dn.net/?f=%3D%5Cint%5Climits%5E%7B9%7D_%7B6%7D%20%7B%5Cfrac%7B1%7D%7B7%7D%5Ccdot%20e%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%20x%7D%7D%20%5C%2C%20dx%20%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B7%7D%5Ccdot%20%5Cint%5Climits%5E%7B9%7D_%7B6%7D%20%7Be%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%20x%7D%7D%20%5C%2C%20dx%20%5C%5C%5C%5C%3D%5B-e%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%20x%7D%5D%5E%7B9%7D_%7B6%7D%5C%5C%5C%5C%3De%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%206%7D-e%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%209%7D%5C%5C%5C%5C%3D0.424373-0.276453%5C%5C%5C%5C%3D0.14792%5C%5C%5C%5C%5Capprox%200.148)
Thus, the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.
Jim’s statement is correct for all numbers greater than or equal to 0. It does not work for negative numbers. For example, the floor of –3.5 is –4, where –4 is not equal to –3, the number before the decimal point. Hope this helps!
Thanks!
-Charlie
The answer is A
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Answer:
This question i think google could help you if not more then brainy if u dont get a answer fast enough
Step-by-step explanation:
hope i helped u a little bit i know i didn't have a answer.