Pl.solve it step by step.?

Question

Suppose we build a model of a production line that produces on the average 6 per cent defective items. Defective items occur randomly. Items are packed for sale in lots of four. We are interested in kn0wing what percentage of the lots contains no defectives. Answer using Monte Carlo simulation technique.(Take simulated sample of 20 lots) Also compare with exact results.

Answer 1

Exact result:

Chance of any single unit being good is 96%. Chance that all four in one lot is 0.96^4 = 85%.

Monte Carlo:

I used XL random number generator.

let numbers = 00 to 99, and let 0 - 3 be fails, all other good.

formulas: =INT(RAND()*100)

=1*OR(B7<4,C7<4,D7<4,E7<4)

This gives 0 if passes, 1 if fails. Sum the fails at the bottom.

Results: Do 20 samples of 4 numbers, count all those that have at least one number from 0 to 3.

20 samples ranged from .7 to 1.0

400 samples ranged from 82% to 89%.

4000 samples was 0.858

A random listing with 4 fails (trials 1,2,18,19), which is 0.8 good:

n   #1   #2   #3   #4

1 99 19 68 1

2 24 90 91 3

3 50 71 46 65

4 77 21 76 99

5 16 46 17 68

6 33 86 7 95

7 87 63 8 28

8 63 63 47 63

9 86 25 39 14

10 90 78 67 62

11 16 78 44 49

12 86 98 97 6

13 77 92 42 43

14 43 50 62 23

15 36 71 59 80

16 95 91 50 51

17 68 99 41 25

18 68 45 2 90

19 0 84 74 96

20 29 48 83 85