00:01
Hi, i'm david and i'm here to help you understand your question.
00:03
Now let me bring up your question here.
00:06
In the question here, we are going to discuss about the normal distribution.
00:12
And let me remind you that if we have the n greater equal to the 30, the sum of mean x -par will be approximate to the normal in such a way that the mean of the x -par equal to the meaning of population, standard division x -par equal to the standard division of population divided base square root of the n.
00:29
And if it follows by the normal, if we turn the x bar minus the mean of a standard division, we will obtain the standard normal.
00:38
Also, if we have the random variable x that will follow by the uniform from a and p, the mean of the x equal to the a plus p divided by 2, and the standard division of the x by the formula equal to the p minus a, the value by 12, square, the top, and under the standard division of the x by the formula, square root.
01:01
Now in this question here we are given the age x will follow by the uniform from the 6 to 11.
01:08
So from here we can find the mean of the x by the formula here it will equal to the 6 plus 11 of 2 it will equal to the 17 different with you equal to the 8 .5 with the standard division by this formula equal to the square root of the 11 minus 6 square divided by 12 and then we have 5 squared to 1 .4 squared equal to the 1 .443.
01:43
And from here we can answer the first part here for the question a, where you will have the distribution of the x bar will follow by the normal, where the mean it will equal to the mean of the proportion will be the 8 .5.
02:01
The standard division, sigma on the x bar, by the formula we need to divide by the square root the n will be the 1 .443 over square root of the n will be n equal to the 45.
02:17
So the 1st square root of 45 we get equal to the 0 .215 and that's going to be the answer from the a.
02:29
For the p we want to find the probability that the x bar will between the 8 and the 8 .3.
02:39
To find this probability i need to convert the xx into the z...