00:01
There is given a normal distribution for this question and the mean denoted by mew that was given as 8 .35 hours daily and the standard division was also given here denoted by sigma and that was given as 2 .5.
00:15
So i can just define random variable x which is normally distributed.
00:19
So the mean is 8 .35 and the standard division which is 2 .5 here.
00:24
So what do we need to get? so for part a, what is the problem that the household was television between? 3 and 12 hours so we need to get the random variable x which is between 3 and 12 to get this problem i'm going to use the graphic display calculator application normal cdf over boundary 3 upper boundary is 12 and the mean is 8 .35 and the standard division which is 2 .5 here let me get the second variance in the normal cdf lower boundary 3 and the upper boundary 12 so the mean is 8 .35 and the standard division which is 2 .5 here so the probability would be 0 .91 and 17.
01:04
And what about for part b? so how many hours of television viewing must be a hold have in order to be top 2 %? let's say x1 be the minimum hour for top 2%.
01:21
Let me graph it.
01:23
Let's say this is the normal distribution we have.
01:26
Top 2 % is this region and this is x1 here.
01:29
This is 2%.
01:30
I can write at 0 .0 .0 .000.
01:31
In order to get the x1 here, we need to just find the area of this yellow shaded region.
01:37
And we know that the total area under the curve, which is 1.
01:40
So the probability of x is less than x1, which is 1 minus 0 .02, which would be 0 .98.
01:46
To get the x1 value here, again, i'm going to use the graphing display calculator application inverse norm, so the area from left to right...