Question

based upon historical data, it is known that 8% of 12 - egg cartons contain at least one broken egg. a grocery store manager would like to carry out a simulation to estimate the number of cartons, in a sample of 10, that would contain at least one broken egg. she assigns the digits to the outcomes:
01 - 08 = carton contains a broken egg
09 - 99, 00 = carton does not contain a broken egg
here is a portion of a random number table.
1 31645 03495 96193 10898
2 67940 85019 98036 98252
3 21805 26727 73239 53929
4 03648 93116 98590 10083
5 71710 46584 35453 98153
in the first trial, line 1, one of the first 10 double - digit numbers is between 01 and 08, meaning that 1 of the 10 cartons of eggs contains at least one broken egg. starting at line 2 and using a new line for each trial, carry out 4 more trials. in what proportion of these 5 trials is there no carton with broken eggs?

Explanation:

Step1: Analyze line 2

Take two - digit numbers from line 2: 67,94,08,50,19,98,03,69,82,52. The numbers between 01 - 08 are 08,03. So there are 2 cartons with broken eggs.

Step2: Analyze line 3

Take two - digit numbers from line 3: 21,80,52,67,27,73,23,95,39,29. There are 0 numbers between 01 - 08.

Step3: Analyze line 4

Take two - digit numbers from line 4: 03,64,89,31,16,98,59,01,00,83. The numbers between 01 - 08 are 03,01. So there are 2 cartons with broken eggs.

Step4: Analyze line 5

Take two - digit numbers from line 5: 71,71,04,65,84,35,45,39,81,53. The number between 01 - 08 is 04. So there is 1 carton with broken eggs.

Step5: Count the number of trials with no broken - egg cartons

We have 5 trials (line1 - line5). From the above analysis, only line3 has 0 cartons with broken eggs.

Answer:

The proportion is $\frac{1}{5}=0.2$