Question

From Allen (1990, 113, 162). The interactive computer system at Gnu Glue has 20 communication lines to the central computer system. The lines operate independently and the probability that any particular line is in use is 0.6. () What is the probability that 10 or more lines are in use? Compare the exact binomial solution and the normal approximation solution to this problem. (ii) Do the assumptions needed for the normal approximation make sense?

Answer 1

Binomial = 0.8724

Normal approximation = 0.873

Explanation:

Given that :

Sample size (n) = 20

p = 0.6

q = 1 - p = 1 - 0.6 = 0.4

.using normal approximation :

Mean = np

Mean (m) = 20 * 0.6 = 12

Standard deviation (s) = √npq

s = √(20 * 0.6 * 0.4) = 2.1908902

Probability of 10 or more lines in use :

p(x ≥ 10)

Applying for correction:

p(x ≥ 10 - 0.5) = p(x ≥ 9.5)

Using the z formula :

Z = (x - m) / s

Z = (9.5 - 12) / 2.1908902

Z = - 1.141

p(Z > - 1.141) = 0.87307 (Z probability calculator)

p(Z > - 1.141) = 0.873

Using the binomial distribution :

P(x = x) = nCx * p^x * (1 - p)^(n-x)

p(x ≥ 10) = p(x = 10) + p(x = 11) +.... + p(x = 20)

Using calculator :

p(x ≥ 10) = 0.87247

= 0.873