00:01
In this problem, we are given that during a national sleep survey, it was observed that the average hours which adults sleep, this is equal to 7 every night and the standard deviation, well that's equal to 1 .5 hours.
00:18
And we have to assume here that the time which the adults sleep, let's represent that with the variable x, this is having normal distribution.
00:28
So first we have to get the percentage of the adults who sleeps between 4 .75 and 9 .25 hours.
00:39
So using this formula, we can get the z -scores corresponding to these two raw scores.
00:44
So this will be the probability that the z -score is greater than 4 .75 minus 7 over 1 .5, but it will be less than 9 .25 minus 7 divided by 1 .5.
00:57
And that will be the probability that the z -score is greater than minus 1 .5 but less than 1 .5.
01:04
And this will be the probability that the z -score is less than 1 .5 minus probability that the z -score is less than minus 1 .5.
01:12
And using the z -table, we get these probabilities as 0 .9332 and 0 .0668 respectively.
01:22
So when we subtract, we will get 0 .8664 as the result.
01:28
And in the next part, we are required to determine the range of sleeping times, which will consist of the middle 97 % of the individuals.
01:40
So first we can convert this in terms of percent.
01:44
So 0 .8664, this will be about 86 .64%.
01:48
And in the next part, we see that the probability that the z -score is greater than minus z -star but less than z -star.
01:58
This is 97%, which will be 0 .97...