4)-In the above example, the Cumulative distribution of the UI.1U) is: .A. X1-0.
ID: 3368665 • Letter: 4
Question
4)-In the above example, the Cumulative distribution of the UI.1U) is: .A. X1-0.1 B. 0.1X .C. 2X-0.1 .D.)0.1x 1/10 .E. None 5)- Given that workers in a company logon t0 the central computer, on average every 4 minutes, if the time between the logons is assumed to be Exponentially distributed, the probability that there is no logon in 6 minutes is .A. 0.323 B. 0.021 .C. It is the same as the average .D. 0.082 .E. None. 6)-Given X N(300, 15), the area to the right of x 330.. .A. Is the same as the area to the right of z 2. .B. PIX> 330) is the same as the area to the right of z 2&to; the left of X - 270. .C. Is the same as the area to the left ofx 270 D. Is the same as the area to the left ofx 330 .E. All of above 7)-Given X N(300, 15), find the Inter-quartile Range (IQR03 Q1) . A. 0.50 B. 310.12 . C. 20.24 D. 289.9 8)-Given X N(300, 10). If X is the dimension of a product you buy from a company who claim the specification is 300 ± 12. , what % of x values are within specification limit. . A. 11.51% B. 76.98% . C. 88 49% D. 10.98% E. None PROBLEM # 2 (15 PC)INTS) Mathermatical tests have shown that the random variable X has the following probability mass function: RX) (14) (X-1), where: 2c- X4 Using this information, answer the following questions: 1). Is fX) is a probability distribution? (prove/disprove. 2). Assuming it is a probability distribution: determine Mean, Standard Deviation, and Cumulative distribution function of X and graph it. 3). Determine Pr (2.00 X 3.5), Pr(X 2.1), and Prix>-2.9) 15Explanation / Answer
Solutions
Q4
If X ~ U(1, 11), its pdf is f(x) = {1/(11 - 1)}dx} = dx/10 and its CDF is F(x) = P(X < x)
= integral[1, x]f(x)dx
= integral[1, x](dx/10)
= [x/10](1, x)
= (x/10) – (1/10)
= 0.1x - (1/10) Option D ANSWER 1
Q5
If X ~ Exponential with parameter ? (average inter-event time), the pdf (probability density function) of X is given by f(x) = (1/?)e-x/?, 0 ? x < ? …………………………(1)
CDF (cumulative distribution function), F(t) = P(X ? t) = 1- e-t/? ………….…(2)
From (2), P(X > t) = e-t/? ………………………….……………………………(3)
Given X = time between logons ~ Exp(4), probability there is no logon in 6 minutes
= P(X > 6), which vide (3) above,
= e-6/4
= e- 1.5
= 0.223 Option E ANSWER 2
Q6
If a random variable X ~ N(µ, ?2), i.e., X has Normal Distribution with mean µ and variance ?2, then,
Z = (X - µ)/? ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)
Thus, given X ~ N(300, 152), area (probability) to the right of x = 330 is the same as
area (probability) to the right of z = {(330 – 300)/15}, i.e., area (probability) to the right of
z = 2 Option A ANSWER 3
DONE
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