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slavikrds [6]
3 years ago
14

The commuter trains on the Red Line for the Regional Transit Authority (RTA) in Cleveland, OH, have a waiting time during peak r

ush hour periods of eight minutes ("2012 annual report," 2012).
a.) Find the height of this uniform distribution.


b.) Find the probability of waiting between four and five minutes.


c.) Find the probability of waiting between three and eight minutes.


d.) Find the probability of waiting five minutes exactly.

Mathematics
1 answer:
eduard3 years ago
8 0

Answer:

a) height of this uniform distribution = 1/8 = 0.125.

b) P( 4 < x < 5) = (5 - 4)*0.125 = 0.125

c) P( 3 < x < 8) = (8 - 3)*0.125 = 0.625

d) P(x = 0) = 0*0.125 = 0

Step-by-step explanation:

Height of this uniform distribution is same as finding the probability for each waiting minutes.

Height of uniform distribution = 1/n. Where n is the number of occurence, in this n = 8.

a) height of this uniform distribution = 1/8 = 0.125.

Let random variable x = waiting time.

b) probability of waiting between four and five minutes

P( 4 < x < 5) = (5 - 4)*0.125 = 0.125

c) probability of waiting between three and eight minutes

P( 3 < x < 8) = (8 - 3)*0.125 = 0.625

d) probability of waiting five minutes exactly.

Since this would be just one line, and the width of the line is 0, then the

P(x = 0) = 0*0.125 = 0

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Answer:

The solution is  \frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] +  C

Step-by-step explanation:

From the question

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The  indefinite integral is  mathematically represented as

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=>   \frac{du}{dx} 2e^{2x}

=>   2 e^{2x}dx =  du

So

\int\limits  {\frac{e^{2x}}{ 25 + e^{4x}}} \, dx =  \int\limits  {\frac{1}{ 2(25 + u^2)} } \, du

= \frac{1}{2} \int\limits  {\frac{1}{ 25 + u^2)} } \, du

=  \frac{1}{2} \int\limits  {\frac{1}{ 5^2 + u^2)} } \, du

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Now substituting for  u

\frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] +  C

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Answer:

z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23    

The p value would be given by:

p_v =2*P(z  

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

Step-by-step explanation:

Information given

\bar X=46.5 represent the mean

\sigma=1.1 represent the population standard deviation

n=150 sample size  

\mu_o =46.7 represent the value to verify

\alpha=0.05 represent the significance level for the hypothesis test.  

z would represent the statistic

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We want to test if the true mean for this case is 46.7, the system of hypothesis would be:  

Null hypothesis:\mu = 46.7  

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Since we know the population deviation the statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Replacing we got:

z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23    

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