00:01
Hi, i'm david and i'm here to help you and change your question.
00:03
In the question here, we are going to discuss about the distribution of the sample mean.
00:07
Let me remind you that if the x follows by the normal, between the mean and the standard deviation, and then the symbol mean xx will also follow by the normal.
00:17
Well, when the mean of the xx equal to the mean in the population, standard division xx equal to the sigma divided by square into the n.
00:24
And because it follows by the normal, if it will then the xx will minus the mean over the standard deviation.
00:30
We have standard standard normal.
00:32
In the question here we have the x bar, when the x follows by the normal, with the mean equal to the 2 .72, and the standard division equals to 0 .03, the n equal to the 10.
00:50
So the first question a here, we will have the sum of mean x by the normal, where we have the mean of the x bar equal to the mean of the population equal to the 2 .72, standard division, x by, equal to the sigma defined square n d 'n.
01:05
Then 1 0 03, the vascular 2, the 10, equal to the 0pon 0905.
01:19
And from here, we see the answer for the question a, it will be, is a personal normally distribution and the symbol distribution of the symbol on the 6 and will be approximately normal, symbol 6 and will be appresumate and that will be the answer for the question d.
01:45
And then for the question b, once you find the probability that the sample mean x bar could be smaller than the 2 .89.
01:54
To find this probability, i need to convert the x bar into the z.
01:58
To do it and need to apply this formula here.
02:02
And then we'll take the 2 .89 will minus the mean will be the 2 .72 over the mean, the standard division.
02:09
Equal to the 0195 and if we compute and have the zee smaller than 2 .89 minus 2 .72 the fundamental standard division equal to the 17 .91 92 and then this probability will be approximately to the 1 and that's will be the answer for the question b.
02:39
This one will be 69 sorry this one will be the 69 not the 89 it doesn't be the 69 here so we have 2 .69 minus 2 .72 divided by the zp .095 equal to the minus 3 point one 6 and every computer and use the z table so if we use the z table we get this one would be minus 3 point 16 then get equal to the 0 .0 008 and that will be the answer for the b for the c i want to find the probability that the x bar will be between the 2 .70 and the 2 .74 again i can convert the x point into the z and then we get equal to 2 .70 minus 2 point 72 over the 0 .0095 2 .7 .74 minus 2 .72 divided by the zpon 095.
03:54
And if we'll compute again, the z will be between 2 .74 minus 2 .72 divided by 0 .0 95 equal to the 2 .105 and this one will be the minus 2 .105 and this one will be the minus 2.
04:13
105 and if we compute this one using the normal distribution can equal to 2 .105, 1 minus 2 .101 .1765 can equal to the 0 .96, 477...