The pdf of a random variable X is given by: -0.2x f(x) = 0.2e for x > 0 Please find P(x < 1). 0.4512 0.3297 0.8187 0.1813
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- Areas under normal distribution curves can be used calculate probabilities of many random variables in the natural and social sciences. The graph of the Standard Normal Curve, with a mean of 0 and a standard deviation of 1 is given below, and its function is given by 1 f(x) = √2T e 1 The integral of e dx would be used to calculate the probability of an event that occurs √2T -1.5 between -1.5 to 1.5 standard deviations from the mean. (-1.5, 0.13) 1 Unfortunately, √2π -1.5 methods to approximate the values. 1.5 -1.5 1.5 (-1, 0.24) Using Simpson's Rule, (-0.5, 0.35) e dx has no proper closed form antiderivative, so we must use numerical 1.5 -1.5 1-0.5 -0-2 Using the Midpoint Rule with three rectangles of equal width, dx 0 Using the Trapezoidal Rule with six trapezoids of equal width, (0, 0.4) 0.5 Round answers to 4 decimal places when appropriate. (0.5, 0.35) (1, 0.24) 1.5 Ibe 1.5 1.5 [ 13 e-t de = \ е 1.5 e-2 е dx (1.5, 0.13)The pdf of a random variable X is given by: f(x) = xln(1.for 4 ≤ x ≤ 6 Please find Var(X). 4.94 24.66 0.5766 0.3325Find the standard deviation. Р(x) | x*Р(x) (x - 4)°P(x) 2 0.1 2(0.1) = 0.2 4 0.3 4(0.3) = 1.2 6 0.4 6(0.4) = 2.4 8 0.2 8(0.2) = 1.6 00
- f(x) = (2x + 1)/c, 0 ≤ x ≤ 4. Find the variance of the distribution. 438/c² 409.2/c² O 431.6/c² O 419.6/c²Suppose X is a random variable defined as final cost of project estimated cost of project which has a PDF as follows: [0 xa a. Determine the value of a. b. What is the probability that the final cost of a project will exceed its estimated cost by 25%? c. Determine the mean value and standard deviation of X.Give the answer for question 4
- 0.3507 x 0The geometric distribution gives the probability that the first success occurs at the xth trial with success probability p. f(x)=(1-p)x-1p, x=1,2,3,... Show that E(X)=1/pSuppose the random variable X has the following PDF: 2 √2π This distribution is often called half-normal. Find E(X) and V(X). f(x) = , for x > 0.3 The median value y of a continuous random variable is that value such that F(y) = 0.5. Find the median value of the random variable in 90.5 = f(y) = 2 25 -Y, 0 ≤ y ≤ 5, elsewhere. 0,Q1:Let X be a random variable with pdf f (x) = cx + d for 0 = a- Find the values of constants c and d. b- Find the cdf of X. c- find P(IX| <). P(PavanRecommended textbooks for youAdvanced Engineering MathematicsAdvanced MathISBN:9780470458365Author:Erwin KreyszigPublisher:Wiley, John & Sons, IncorporatedNumerical Methods for EngineersAdvanced MathISBN:9780073397924Author:Steven C. Chapra Dr., Raymond P. CanalePublisher:McGraw-Hill EducationIntroductory Mathematics for Engineering Applicat…Advanced MathISBN:9781118141809Author:Nathan KlingbeilPublisher:WILEYMathematics For Machine TechnologyAdvanced MathISBN:9781337798310Author:Peterson, John.Publisher:Cengage Learning,Advanced Engineering MathematicsAdvanced MathISBN:9780470458365Author:Erwin KreyszigPublisher:Wiley, John & Sons, IncorporatedNumerical Methods for EngineersAdvanced MathISBN:9780073397924Author:Steven C. Chapra Dr., Raymond P. CanalePublisher:McGraw-Hill EducationIntroductory Mathematics for Engineering Applicat…Advanced MathISBN:9781118141809Author:Nathan KlingbeilPublisher:WILEYMathematics For Machine TechnologyAdvanced MathISBN:9781337798310Author:Peterson, John.Publisher:Cengage Learning,