Exam 2 Key MA 223 1. (a) P (A) = 0.0.88 (that is 88 of 100 bearings have acceptable roundness), while P (B) = 0.85. Also P (A|B) = P (A ∩ B) 78/100 = ≈ 0.918 P (B) 85/100 P (B|A) = 78/100 P (A ∩ B) = ≈ 0.886 P (A) 88/100 and (b) If A and B are independent then P (A∩B) = P (A)P (B), or 0.78 = (0.88)(0.85) = 0.748, which is not true. So these are NOT independent. Alternatively, simply note that P (A) ̸= P (A|B), and/or P (B) ̸= P (B|A). If the bearing passes roundness it is more likely to pass the average diameter test (and vice-versa). 2. (a) E(W ) = E(X) − 2E(Y ) = −8.0 and V (W ) = V (X) + 4V (Y ) = 16.0. (b) E(x̄) = E(X) = 2.0 and V (x̄) = V (X)/n = 4.0/n. (c) If n = 100 then x̄ is approximately normal, Z = (x̄ − 2)/(1/5) is standard normal, so P (x̄ ≤ 1.6) = P (Z ≤ −2) ≈ 0.0228. from the table. √ √ 3. (a) The interval is 4.94 − (1.96)(0.1)/ 30 < µ < 4.94 + (1.96)(0.1)/ 30 or 4.904 < µ < 4.976. √ √ (b) The confidence interval width is given by q = (1.96)(0.1)/ n, so solve (1.96)(0.1)/ n ≤ 0.02 for n to find n ≥ 96, roughly. √ 4. We have x̄ = 1200 and s = 150; the statistic t = (x̄−µ)/(s/ n) follows a t distribution with 15 df, so from the table we have −1.753 < (1200 − µ)/37.5 < 1.753, leading to 1134 < µ < 1266. 1