1. A mail-order computer business has six telephone lines. Let X ...

P [X ≤ 2] =
=
10
x=5
10
x
(0.5) x [1 − 0.5] 10−x
10!
5!(10 − 5)! [0.000976562]
10!
+
6!(10 − 6)! [0.000976562]
10!
+
7!(10 − 7)! [0.000976252]
=
10!
8!(10 − 8)! [0.000976562]
10!
+
9!(10 − 9)! [0.000976562]
10!
+
10!(10 − 10)! [0.000976252]
= (252 + 210 + 120 + 45 + 10 + 1) × 0.000976562
(c) Determine P [1 ≤ X ≤ 4].
Ans Required probability:
= 0.6230469
(1 + 10 + 45 + 120 + 210) × 0.000976562
= 386 × 0.000976562
= 0.3769531
Note that we could also have arrived at the same answer, without
computing binomial coefficients, by observing that,
P [1 ≤ X ≤ 4] = P [X = 0] + [1 − P [X ≥ 5]]
= 1 + 0.000976562 − 0.6230469 = 0.3769
(d) What is the probability that none of the 10 boards are defective?
Ans This is the probability
P [X = 0] = 0.0009765
(e) Calculate the expected value and standard deviation of X.
Ans We know that any Binomial r.v. can be defined as the
sum of independent Bernoulli trials, i.e., if X ∼ BIN(n, p), then
4