2. 1/4 points | Previous Answers DevoreStat9 12.E.008. My Notes Ask Your Teacher

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2. 1/4 points | Previous Answers DevoreStat9 12.E.008. My Notes Ask Your Teacher An article considered regressing y=28-day standard-cured strength (psi) against x- accelerated strength (psi). Suppose the equation of the true regression line is y-1900+ 1.3x, and that the standard deviation of the random deviation c is 350 psi (a) What is the probability that the observed value of 28-day strength will exceed 5000 psi when the value of accelerated strength is 2100? (Round your answer to four decimal places.) 0.1469 (b) What is the probability that the observed value of 28-day strength will exceed 5000 psi when the value of accelerated strength is 2600? (Round your answer to four decimal places.) 0.1335 X (e) Consider making two independent observations on 28-day strength, the first for an accelerated strength of 2100 and the second for x 2600. What is the probability that the second observation will exceed the first by more than 1000 psi? (Round your answer to four decimal places.) .2799 X (d) Let Yi and denote observations on 28-day strength when x-x, and x-x2, respectively. By how much wouldx, have to exceed x, in order that PY2 .95 (Round your answer to two decimal places.) You may need to use the appropriate table in the Appendix of Tables to answer this question. Need Help? Talk to Tutor

Explanation / Answer

b) when x = 2600

y^ = 1900+1.3*2600 = 5280

mean = 5280 , sd = 350

P(Y > 5000)

=P(Z >( 5000 -5280)/350)

= P (Z>0.8)=0.7881

c)

Y1 - N(4630,350^2)

Y2 - N( 5280,350^2)

P(Y2 - Y1 > 1000)

Y2 - Y1   - N(5280 -4630 , 350^2 + 350^2)

= N(650,494.97474^2)

P(Y2 - Y1 > 1000)

= P(Z > (1000 - 650)/494.9747)

= P (Z>0.71)=0.2389

d)

P(Y2 > Y1) = 0.95

P(Y2 - Y1 > 0) =0.95

P(Z> z*) = 0.95

z* = -1.645

Y2 - Y1 = N(1.3*(x2-x1), 494.97474^2)

hence

-1.645 = (0 -1.3*(x2-x1) )/494.97474

hence

x2-x1 = 494.97474 *1.645/1.3

= 626.3334

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