Question 1204105
the probability thqt the whole shipment will be accepted is equal to 0.993131178 this is approximately 99.31% when the percent is rounded to 2 decimal places.
that's the probability that less than or equal to 3 batteries in a sample of 38 batteries will be defective.
the probability that more than 3 batteries will be rejected in a random sample of 38 is equal to 100% - 99.31% = .69%.


i considered this a binomial probability problmm.
the calculations were done in excel.


the formula for each probability is p(x) = p^x * q^(n-x) * c(n,x)
for example, the probability that 3 will be rejected is:
p(3) = .02^3 * .98^(38-3) * c(38,3) = 0.03327662.
that's the same number for p(3) in the spreadsheet.
c(n,x) = n! / (x! * (n-x)!).
that becomes c(38,3) = 38! / (3! * (38-3)!) which is equal to 8436.
this is the same number in the spreadsheet for p(3).


here's the spreadsheet with all the calculations.


<img src = "http://theo.x10hosting.com/2023/100701.jpg">


<img src = "http://theo.x10hosting.com/2023/100702.jpg">