00:01
Let us work on this problem.
00:02
A grocery chain offers discount rates.
00:04
We can start with a given information.
00:06
We are given with probability p equals 5 percentage.
00:10
You can write this 5 percentage as a decimal 5 over 100, which is 0 .05.
00:18
We use binomial distribution to find out mean and standard deviation.
00:24
The formula to find out mean here is n into p and standard deviation is square.
00:30
Root of n p into 1 minus b.
00:34
We have the sample size n equals 300.
00:37
Let us substitute this and find out our mean.
00:41
So mean becomes 300 into 0 .05 which is equals 50.
00:49
And standard deviation substitute the values.
00:52
We have square root of 300 into 0 .05 into 1 minus 0 .05.
01:00
So the standard deviation is 3 .7749.
01:07
Let us go to part a.
01:09
Part a says, company has 300 employees, what is the probability that company a will have at least 30? so when we have at least 30, we need to find out probability of x greater than are equal to 30.
01:27
Let us use the correction factor.
01:29
When we use the correction factor we need to use probability of x gradethna are equal to x minus 0 .5.
01:37
Let us subtract 0 .5 from this.
01:40
We get probability of x greater than are equal to 29 .5.
01:45
We can use mean and standard deviation to find out the z score.
01:51
So we have probability of z greater than are equal to 30 minus mean.
01:58
Mean is our 15 over standard deviation is 3 .7749.
02:06
Let us calculate this.
02:08
We get probability of z greater than are equal to 3 .84.
02:17
Now we can use our z table to find out the answer...