Question

The downtime per day for a computing facility has mean 4 hours and standard deviation .8 hour. Suppose that we want to compute probabilities about the average daily downtime for a period of 30 days.
i. What assumptions must be true to use the result of Theorem 7.4 to obtain a valid approximation for probabilities about the average daily downtime?
ii. Under the assumptions described in part (i), what is the approximate probability that the average daily downtime for a period of 30 days is between 1 and 5 hours?

Answer 1

Answer:

1

Step-by-step explanation:

Given :

Mean, μ = 4

Standard deviation, s = 0.8

Sample size, n = 30

The distribution is independent.

Z = (x - μ) / s /sqrt(n)

Probability that downtime period is between 1 and 5

P(1≤ x ≤ 5) :

[(x - μ) / (s /sqrt(n))] ≤ Z ≤ [(x - μ) / (s /sqrt(n))]

[(1 - 4) / (0.8 /sqrt(30))] ≤ Z ≤ [(5 - 4) / (0.8 /sqrt(30)]

[-3 / 0.1460593] ≤ Z ≤] 1 / 0.1460593]

P(-20.539602 ≤ Z ≤ 6.8465342)

P(Z ≤ 6.8465342) - P(Z ≤ - 20.5396)

P(Z ≤ 6.8465342) = 1 (Z probability calculator)

P(Z ≤ - 20.5396) = 0 (Z probability calculator)

1 - 0 = 1