00:02
We derive this data from the introduction of the problem.
00:05
We have a sample space of 120 rooms and a p or a success rate of 0 .75.
00:13
For the first question, we are to find the probability that at least half of the rooms are occupied on any given day.
00:23
And that is equal to the probability that x is greater than half of the rooms that is 120 divided by 2.
00:33
So we have x to be greater than or equal to 60.
00:39
This problem follows a binomial distribution, but since the n is very large, i'm going to approximate it using the normal distribution.
00:50
Therefore, we need to find the mean and the standard deviation.
00:58
Now, the mean is giving us n times p.
01:02
Therefore, we have 120 times 0 .75, and this is equal to 90.
01:11
The standard deviation is equal to square root of n times p multiplied by 1 minus p.
01:29
So we have 120 times 0 .75 multiplied by 1 minus 0 .75.
01:44
We have the square root of 22 .5 which is equal to 4 .74.
01:59
Now now we have our mean and standard deviation.
02:03
Head to use a normal approximation to the binomial distribution to compute the probability.
02:14
To do that, we need to find the z core corresponding to this value.
02:20
The formula is given as z is equal to x minus the mean divided by standard deviation.
02:26
So we have the probability that z is greater than or equal to the x value which is 60 minus the mean 90 divided by the standard deviation 4 .74.
02:52
So i forgot to include this.
02:55
When we are approximating from the binomial distribution to the normal distribution, since it is from a discrete to a continuous probability, we use what we call continuity and by continuity 60 becomes 60 minus 0 .5.
03:29
Which is equal to 59 .5...