129) Find the probability that in 200 tosses of a fair die, we will obtain at least 30 fives

This is the concept of probability and statistics. To find the probability of obtaining at least 30 fives that in 200 tosses of a fair die we proceed as follows;

Probability space of a fair die is x={1,2,3,4,5,6}

probability of obtaining a 5 is P (x) = 1/6, the trials are independent.

number of trials is n=200

This gives us a binomial distribution, you can find this probability exactly by finding the probability if getting exactly 30 fives plus 31 fives plus 32 fives, ... 200.

However, since the expectation, np=200 (1/6) and n (1-p) = 200 (5/6) = 166.7 are both greater than or equal to 5, this implies that the sample is large enough to use the normal approximation.

The mean of the normal would be np=33.3

The standard deviation (sd) if this normal=sqrt (np (1-p)) = sqrt (200 (1/6) (5/6)) = 5.27

If we were to look at the rectangle over 30, it would start at 29.5 and end at 30.5. Since "at least 30" means that we should include 30. Usin the normal curve;

z = (29.5-33.33) / 5.27=-0.73

Looking at the table:

Ф (-0.73) = 0.7673

Therefore our answer is 0.7673