a.
Obtain
a.
Answer to Problem 113SE
The expression for
The expression for
Explanation of Solution
Given info:
The fusion of two exponential distributions for the modeling behavior is termed as “criticality level of the situation” X by the authors.
The probability density
Calculation:
The mean of the exponential distribution is:
The variance of the exponential distribution is:
The expected mean is obtained as given below:
Thus, the expected mean for mixed exponential distributions is:
The variance of a continuous random variable is given as
Here,
Thus, the variance is:
Thus, the
b.
Find the cumulative distribution of X.
b.
Answer to Problem 113SE
The cumulative distribution function of X is:
Explanation of Solution
Denote the cumulative distribution function (cdf) of X as
For
Since X takes the positive values from the interval
For
Hence, the cumulative distribution function of X is:
c.
Find the value of
c.
Answer to Problem 113SE
The value of
Explanation of Solution
Calculation:
The value of
Substitute
Thus, the value of
d.
Find the probability that the X is within one standard deviation from its mean value.
d.
Answer to Problem 113SE
The probability that the X is within one standard deviation from its mean value is 0.879.
Explanation of Solution
Calculation:
From part (a), the expected mean for two exponential distributions is:
Substitute
The standard deviation of X is given by:
The probability that the X is within one standard deviation from its mean value is obtained as shown below:
Here, the random variable X do not takes negative value.
Thus, the probability that the X is within one standard deviation from its mean value is 0.879.
e.
Find the coefficient of variation for an exponential random variable.
Explain about the value of coefficient of variation for a hyper exponential distribution.
e.
Answer to Problem 113SE
The coefficient of variation for an exponential random variable is 1.
Explanation of Solution
Calculation:
It is given that the Coefficient of variation for a random variable X is:
From part (a),
On simplifying,
The coefficient of variation for random variable X in hyper exponential distribution is:
Denote
Then,
The algebraic expression shows that the value of
Thus, the value of coefficient of variation is greater than 1.
f.
Find the coefficient of variation for an Erlang distribution.
f.
Answer to Problem 113SE
The coefficient of variation for an Erlang distribution is
Explanation of Solution
Calculation:
The expected mean for the Erlang distribution is:
The standard deviation for the Erlang distribution is:
The coefficient of variation for an Erlang distribution is given by:
Thus, the coefficient of variation for the an Erlang distribution is
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Chapter 4 Solutions
EBK PROBABILITY AND STATISTICS FOR ENGI
- For the following table of data. x 1 2 3 4 5 6 7 8 9 10 y 0 0.5 1 2 2.5 3 3 4 4.5 5 a. draw a scatterplot. b. calculate the correlation coefficient. c. calculate the least squares line and graph it on the scatterplot. d. predict the y value when x is 11.arrow_forward4.1.8 An article in Electric Power Systems Research ["Model- ing Real-Time Balancing Power Demands in Wind Power Sys- tems Using Stochastic Differential Equations" (2010, Vol. 80(8), pp. 966-974)] considered a new probabilistic model to balance power demand with large amounts of wind power. In this model, the power loss from shutdowns is assumed to have a triangular distribution with probability density function f(x) = -5.56 × 10-4 +5.56 × 10-6x, 4.44 x 10-³-4.44 × 10-6x, 0, { x = [100, 500] x € [500, 1000] otherwise Determine the following: a. P(X 800) d. Value exceeded with probability 0.1.arrow_forwardThe true data-generating mechanism for the simulated data is Y = 4Z – 0.8X1 - 3X2 + 6X4 +X5 + error. d.) Note that the outcome of interest Y is a function of X5, and thus balance on X5 is critical for valid causal inference. If units are matched or partitioned into subgroups according to their true propensity scores, are these subgroups expected to be balanced, on average, in terms of X5? Why or why not? Choose the best answer. Yes, because X5 is unrelated to the true propensities. Yes, because the outcome Y is still a function of X5 No, because it is impossible to know if the subgroups are balanced in terms of X5 No, because X5 is independent of all other covariates No, because X5 is not a confounding variable Now fit a single propensity score model that is overspecified. use all five covariates as well as interactions between x3 and each of the other four covariates to predict the propensities. Again, use the fitted model to predict point estimates of the propensity scores for each…arrow_forward
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- A possible important environmental determinant of lung function in children is the amount of cigarette smoking in the home. Suppose this question is studied by selecting two groups: Group 1 consists of 23 nonsmoking children 5−9 years of age, both of whose parents smoke, who have a mean forced expiratory volume (FEV) of 2.1 L and a standard deviation of 0.7 L; group 2 consists of 20 nonsmoking children of comparable age, neither of whose parents smoke, who have a mean FEV of 2.3 L and a standard deviation of 0.4 L. A) Assuming this is regarded as a pilot study, how many children are needed in each group (assuming equal numbers in each group) to have a 95% chance of detecting a significant difference using a two-sided test with α = .05? B) Answer the question in Problem A if the investigators use a one-sided rather than a two-sided test. Suppose 40 children, both of whose parents smoke, and 50 children, neither of whose parents smoke, are recruited for the study. C) How much power would…arrow_forwardA possible important environmental determinant of lung function in children is the amount of cigarette smoking in the home. Suppose this question is studied by selecting two groups: Group 1 consists of 23 nonsmoking children 5−9 years of age, both of whose parents smoke, who have a mean forced expiratory volume (FEV) of 2.1 L and a standard deviation of 0.7 L; group 2 consists of 20 nonsmoking children of comparable age, neither of whose parents smoke, who have a mean FEV of 2.3 L and a standard deviation of 0.4 L. Suppose 40 children, both of whose parents smoke, and50 children, neither of whose parents smoke, are recruitedfor the study.*8.37 How much power would such a study have using atwo-sided test with significance level = .05, assuming thatthe estimates of the population parameters in the pilot studyare correct?*8.38 Answer Problem 8.37 assuming a one-sided ratherthan a two-sided test is used.arrow_forwardIf a random variable X has the moment generating function Mx (t)= 2 - ť Determine the variance of X.arrow_forward
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