00:01
So in this question, basically here we are assuming that the probability of success in the population is 0 .2.
00:07
We have a sample of a100.
00:09
And using this, we want to find in icon a, what is the probability that the proportion will be greater than 0 .16.
00:19
So with these two informations, we can define that the distribution of this proportion is normal, with mean equals to the true proportion, and the standard deviation equals to the square root of the true proportion, times 1 minus the same value, divided by the sample size.
00:35
So now because we have normality here, we can apply the z -score to find this probability.
00:40
And in this case, we basically change here the distribution to be the standard normal distribution.
00:45
But to do that, we need to compute the z -score for this value, which is the value minus the mean of the sample proportion distribution, divided by the standard deviation of the distribution.
00:58
So this here will give us the z -score equals to negative 141...