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.
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