00:01
Hello, so here we're given the mean mu is 100, and the standard deviation sigma is 15, and the iq scores are normally distributed.
00:11
So for part a here, we want the cutoff score above which lies the top 2 % of the distributions.
00:18
We find the z score for the top 2%.
00:21
So that's going to be the probability that z is equal to 0 .02, giving us that the probability, z less than or equal to z is going to be equal to 0 .98.
00:36
We can use a z table here or a calculator to find our z score to be approximately 2 .05.
00:45
And then we convert to an i's q score, which we have x is equal to mu plus z times sigma.
00:51
So 100 plus 205 or 2 .05 times 15.
00:57
And that gives us the minimum iq here to qualify is going to be approximately 131.
01:05
And then the probability for part b we want to compute the probability that x bar greater than equal to 131.
01:15
So that's going to be equal to probability that z is greater than equal to 131 minus 100 over 7 .5...