Problem 7: The number of breakdowns for a university computer system is closely
ID: 3074746 • Letter: P
Question
Problem 7: The number of breakdowns for a university computer system is closely monitored by the director of the computing center, since it is critical to the efficient operation of the center. The number of breakdowns averages 4 per week, with a standard deviation of 0.8 per week. a) Find an interval that must include at least 90% of the weekly figures on number of breakdown. b) The center director promises that the number of breakdowns will rarely exceed 8 in a one-week period. Is the director safe in making his claim? Explain?Explanation / Answer
Solution
Let X = Number of breakdowns per week.
We will assume X ~ N(µ, 2)
We are given µ = 4 and = 0.8 ……………………………………………………………………………(A)
Back-up Theory
If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then,
Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ……………………………………………………………………..……………………..(1)
P(X or t) = P[{(X - µ)/} or {(t - µ)/}] = P[Z or {(t - µ)/}] .………………………………………………………………………………(2)
Probability values for the Standard Normal Variable, Z, can be directly read off from Standard Normal Tables …………..(2a)
or can be found using Excel Function NORMDIST(x,Mean,Standard_dev,Cumulative)……………………………..…………………..(2b)
Property of symmetry: P(Z > t) = P(Z < - t)…………………………………………………………………………………………………………………..….(3)
Now to work out the solution,
Part (a)
Let (t1, t2) be the interval that includes at least 90% of the weekly figures on number of breakdowns. Then, we should have:
P(t1 X t2) 0.9 or {P(X t2) + P(X t1)} 0.1
We will assume that the interval is equal two-tail one, i.e., P(X t2) = P(X t1) and each is 0.05 [This assumption is tenable vide (3) above]
Again, vide (3) above , t1 = - t2…………………………………………………………………………………………………………………………………………(4)
Now, P(X t2) 0.05 => vide (2) and (A) above, P[Z {(t2 - 4)/0.8}] 0.05
=> {(t2 - 4)/0.8} = 1.645 [vide (2b) above]
=> t2 = 5.3160
Also, vide (4) above, => {(t2 - 4)/0.8} = - 1.645 or
t1 = 2.6840
Thus, the desired interval is: [2.684, 5.316] ANSWER
Part (b)
Probability the weekly figures on number of breakdowns will exceed 8 = P(X > 8)
= P[Z > {(8 - 4)/0.8}] [vide (2) and (A) above]
= P(Z > 5)
= 0.000000287 [vide (2b) above]
This probability being very small, there is hardly any chance that the weekly figures on number of breakdowns will exceed 8.
This in turn implies that the director is safe in making the claim.