chapter 1

Chapter 4: Continuous Random Variables and Probability Distributionsc. We will use the conversion 1 lb = 454 g, then 7 lbs = 3178 grams, and we wish to findP⎛⎜⎝3178 − 3432 ⎞⎟ =482 ⎠( x > 3178 ) = P Z >1 − Φ(−.53)= . 7019d. We need the top .0005 and the bottom .0005 of the distribution. Using the Z table, both.9995 and .0005 have multiple z values, so we will use a middle value, ±3.295. Then3432±(482)3.295 = 1844 and 5020, or the most extreme .1% of all birth weights are lessthan 1844 g and more than 5020 g.e. Converting to lbs yields mean 7.5595 and s.d. 1.0608. Then⎛⎜⎝7 − 7.5595 ⎞⎟ =1.0608 ⎠( x > 7 ) = P Z >1 − Φ(−.53)= . 7019P This yields the sameanswer as in part c.46. We use a Normal approximation to the Binomial distribution: X ∼ b(x;1000,.03) ˜N(30,5.394)39.5 − 30P x ≥ 40 = 1 − P x ≤ 39 = 1−P⎜Z ≤⎝ 5.394= 1 − Φ(1.76)= 1 −.9608= .0392⎛⎞a. ( ) ( ) ⎟⎠⎛⎜⎝50.5 − 30 ⎞⎟5.394 ⎠b. 5% of 1000 = 50: P ( x ≤ 50) = P Z ≤ = Φ(3.80)≈ 1. 0047. P( |X - µ | ≥ σ ) = P( X ≤ µ - σ or X ≥ µ + σ )= 1 – P(µ - σ ≤ X ≤ µ + σ) = 1 – P(-1 ≤ Z ≤ 1) = .3174Similarly, P( |X - µ | ≥ 2σ ) = 1 – P(-2 ≤ Z ≤ 2) = .0456And P( |X - µ | ≥ 3σ ) = 1 – P(-3 ≤ Z ≤ 3) = .002648.a. P(20 - .5 ≤ X ≤ 30 + .5) = P(19.5 ≤ X ≤ 30.5) = P(-1.1 ≤ Z ≤ 1.1) = .7286b. P(at most 30) = P(X ≤ 30 + .5) = P(Z ≤ 1.1) = .8643.P(less than 30) = P(X < 30 - .5) = P(Z < .9) = .8159146