1. A mail-order computer business has six telephone lines. Let X ...
1. A mail-order computer business has six telephone lines. Let X ...
1. A mail-order computer business has six telephone lines. Let X ...
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E[X] = (1 × 0.2) + (2 × 0.4) + (3 × 0.3) + (4 × 0.1)<br />
= 0.2 + 0.8 + 0.9 + 0.4 = 2.3<br />
V [X] = E[X 2 ]−(E[X]) 2 = (1×0.2)+(4×0.4)+(9×0.3)+(16×0.1)−2.3 2<br />
= (0.2 + <strong>1.</strong>6 + 2.7 + <strong>1.</strong>6) − 2.3 2 = 0.81<br />
The required r.v. is Y = 100 − 5X, therefore,<br />
E[Y ] = 100 − 5 × E[X] = 100 − 5 × 2.3 = 100 − 1<strong>1.</strong>5 = 88.5<br />
V [Y ] = V [100 − 5 × V [X]] = (5 2 ) × V [X] = 20.25<br />
5. When circuit boards used in the manufacture of compact disc players<br />
are tested, the long-run percentage of defectives is 5%. <strong>Let</strong> X = the<br />
number of defective boards in a random sample of size n = 25, so<br />
X ∼ BIN(25, 0.05).<br />
(a) Determine P (X ≤ 2)<br />
(b) Determine P (X ≥ 5)<br />
(c) Determine P (1 ≤ X ≤ 4)<br />
(d) What is the probability that none of the 25 boards are defective?<br />
(e) Calculate the expected value and standard deviation of X.<br />
(a) Determine P [X ≤ 2].<br />
=<br />
10<br />
0<br />
P [X ≤ 2] =<br />
<br />
(0.5) 0 0.5 10 +<br />
=<br />
(b) Determine P [X ≥ 5].<br />
2<br />
x=0<br />
10<br />
x<br />
10<br />
1<br />
<br />
(0.5) x [1 − 0.5] 10−x<br />
<br />
(0.5) 1 0.5 9 <br />
10<br />
+<br />
2<br />
10!<br />
0!(10 − 0)! [0.000976562]<br />
10!<br />
+<br />
1!(10 − 1)! [0.000976562]<br />
10!<br />
+<br />
2!(10 − 2)! [0.000976252]<br />
= (1 + 10 + 45) × 0.000976562<br />
= 56 × 0.000976562 = 0.0546875<br />
3<br />
<br />
(0.5) 2 0.5 8
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