Answer: 63 defective widgets
Step-by-step explanation:
Given that the proportion should not exceed 5%, that is:
p< or = 5%.
So we take p = 5% = 0.05
q = 1 - 0.05 = 0.095
Where q is the proportion of non-defective
We need to calculate the standard error (standard deviation)
S = √pq/n
Where n = 1024
S = √(0.05 × 0.095)/1024
S = 0.00681
Since production is to maximize profit(profitable), we need to minimize the number of defective items. So we find the limit of defective product to make this possible using the Upper Class Limit.
UCL = p + Za/2(n-1) × S
Where a is alpha of confidence interval = 100 -90 = 10%
a/2 = 5% = 0.05
UCL = p + Z (0.05) × 0.00681
Z(0.05) is read on the t-distribution table at (n-1) degree of freedom, which is at infinity since 1023 = n-1 is large.
Z a/2 = 1.64
UCL = 0.05 + 1.64 × 0.00681
UCL = 0.0612
Since the UCL in this case is a measure of proportion of defective widgets
Maximum defective widgets = 0.0612 ×1024 = 63
Alternatively
UCL = p + 3√pq/n
= 0.05 + 3(0.00681)
= 0.05 + 0.02043 = 0.07043
UCL =0.07043
Max. Number of widgets = 0.07043 × 1024
= 72
