In: Statistics and Probability
In the book Advanced Managerial Accounting, Robert P.
Magee discusses monitoring cost variances. A cost variance
is the difference between a budgeted cost and an actual cost. Magee
describes the following situation:
Michael Bitner has responsibility for control of two
manufacturing processes. Every week he receives a cost variance
report for each of the two processes, broken down by labor costs,
materials costs, and so on. One of the two processes, which we'll
call process A , involves a stable, easily controlled
production process with a little fluctuation in variances. Process
B involves more random events: the equipment is more
sensitive and prone to breakdown, the raw material prices fluctuate
more, and so on.
"It seems like I'm spending more
of my time with process B than with process A,"
says Michael Bitner. "Yet I know that the probability of an
inefficiency developing and the expected costs of inefficiencies
are the same for the two processes. It's just the magnitude of
random fluctuations that differs between the two, as you can see in
the information below."
"At present, I investigate
variances if they exceed $2,659, regardless of whether it was
process A or B. I suspect that such a policy is
not the most efficient. I should probably set a higher limit for
process B."
The means and standard deviations of the cost variances of
processes A and B, when these processes are in
control, are as follows: (Round your z value to 2 decimal
places and final answers to 4 decimal places.):
Process A | Process B | |
Mean cost variance (in control) | $ 3 | $ 1 |
Standard deviation of cost variance (in control) | $5,473 | $9,743 |
Furthermore, the means and standard deviations of the cost
variances of processes A and B, when these
processes are out of control, are as follows:
Process A | Process B | |
Mean cost variance (out of control) | $7,651 | $ 6,169 |
Standard deviation of cost variance (out of control) | $5,473 | $9,743 |
(a) Recall that the current policy is to investigate a cost variance if it exceeds $2,659 for either process. Assume that cost variances are normally distributed and that both Process A and Process B cost variances are in control. Find the probability that a cost variance for Process A will be investigated. Find the probability that a cost variance for Process B will be investigated. Which in-control process will be investigated more often.
Process A | ||
Process B | ||
(Click to select)Process AProcess B is investigated more often
(b) Assume that cost variances are normally
distributed and that both Process A and Process B
cost variances are out of control. Find the probability that a cost
variance for Process A will be investigated. Find the
probability that a cost variance for Process B will be
investigated. Which out-of-control process will be investigated
more often.
Process A | ||
Process B | ||
(Click to select)Process BProcess A is investigated more often.
(c) If both Processes A and B
are almost always in control, which process will be investigated
more often.
(Click to select)Process AProcess B will be investigated more
often.
(d) Suppose that we wish to reduce the probability
that Process B will be investigated (when it is in
control) to .3121. What cost variance investigation policy should
be used? That is, how large a cost variance should trigger an
investigation? Using this new policy, what is the probability that
an out-of-control cost variance for Process B will be
investigated?
k | |
P(x > 4,775) | |
(a)
Probability that a cost variance for Process A will be investigated
= P(A > 2659)
= P(Z > (2659 - 3)/5473) = P[Z > 0.49] = 0.3121
Probability that a cost variance for Process B will be
investigated = P(B > 2659)
= P(Z > (2659 - 1)/9743) = P[Z > 0.27] = 0.3936
Since P(A) < P(B), Process B is investigated more often
(b)
Probability that a cost variance for Process A will be investigated
= P(A > 2659)
= P(Z > (2659 - 7651)/5473) = P[Z > -0.91] = 0.8186
Probability that a cost variance for Process B will be
investigated = P(B > 2659)
= P(Z > (2659 - 6169)/9743) = P[Z > -0.36] = 0.6406
Since P(A) > P(B), Process A is investigated more often
(c)
If both Processes A and B are almost always in control, which
process will be investigated more often.
From part (a)
Process B will be investigated more often
(d)
Probability that a cost variance for Process B will be investigated
= P(B > k) = 0.3121
P(Z > (k - 1)/9743) = 0.3121
(k - 1)/9743 = 0.4899 (Using Standard Normal distribution
tables)
k = 1 + 9743 * 0.4899 = 4774.096
k = 4775 (Rounding to next integer)
Probability that an out-of-control cost variance for Process B
will be investigated = P(B > 4775)
= P(Z > (4775 - 6169)/9743) = P[Z > -0.14] = 0.5557