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Bumek [7]
3 years ago
12

A team of researchers published an article on the study of how vehicles are dispatched based on an airport-based taxi service. T

he researchers modeled this system with an underlying assumption that travel times of successive trips to and from the terminal are independent exponentially distributed random variables with β = 15 minutes. (a) Find the mean and standard deviation of trip time distribution (b) How likely is it for a particular trip to take more than 25 minutes? (c) If two taxis are dispatched together, what is the probability that both of them will be gone for more than 25 minutes? (d) what is the likelihood of at least of one of the taxis returning within 25?
Mathematics
1 answer:
nikklg [1K]3 years ago
3 0

Answer:

a. The mean would be 0.067

The standard deviation would be 0.285

b. Would be of 1-e∧-375

c. The probability that both of them will be gone for more than 25 minutes is 1-e∧-187.5

d. The likelihood of at least of one of the taxis returning within 25 is 1-e∧-375

Step-by-step explanation:

a. According to the given data the mean and the standard deviation would be as follows:

mean=1/β=1/15=0.0666=0.067

standard deviation=√1/15=√0.067=0.285

b. To calculate How likely is it for a particular trip to take more than 25 minutes we would calculate the following:

p(x>25)=1-p(x≤25)

since f(x)=p(x≤x)=1-e∧-βx

p(x>25)=1-p(x≤25)=1-e∧-15x25=1-e∧-375

c. p(x>25/2)=1-p(x≤25/2)=1-e∧-15x25/2=1-e∧-187.5

d. p(x≥25)=1-e∧-15x25=1-e∧-375

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A survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10. (a)
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a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b) The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

<u>Step-by-step explanation</u>:

Step:-(i)

Given data a survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10

The sample size 'n' = 25

The mean of the sample   x⁻  = $28

The standard deviation of the sample (S) = $10.

Level of significance ∝=0.05

The degrees of freedom γ =n-1 =25-1=24

tabulated value t₀.₀₅ = 2.064

<u>Step 2:-</u>

The 95% of confidence intervals for the average spending

((x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )

(28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )

( 28 - 4.128 , 28 + 4.128)

(23.872 , 32.128)

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b)

Null hypothesis: H₀:μ<30

Alternative Hypothesis: H₁: μ>30

level of significance ∝ = 0.05

The test statistic

t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }

t = \frac{28-30 }{\frac{10}{\sqrt{25} } }

t = |-1|

The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

<u>Conclusion</u>:-

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

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