00:01
There is given a normal distribution and the mean value for the population, which is denoted by the symbol mu, that is given as 16, and the standard deviation, which is given us 6 .5.
00:13
Let's say x be the random variable for this normal distribution, that is 16 and 6 .5.
00:19
So the answer for part a, what is the distribution for x, which is, i mean, normal distribution, which is 16.
00:30
6 .5 so this is the answer for part a so and also 34 random and select people so there is given sample size which is 34 and what is the distribution for this one so let's say x -4 is the normal distribution for the sample for the sample in order to get the sample standard division we have the formula this one the population mean divided by the sample size vertical sample size so the standard division for x which is this is 6 .5 divided by radical 34 which is 6 .5 divided by this is second and radical 34 which gives us the value of 1 .1.
01:18
This is 115.
01:21
So for this small places we have to give the answer with four decimal places which is 47.
01:30
So the normal distribution x bar has the variable so the mean value for the sample which is same with the population mean which is 16 and the standard deviation for the sample which is 1 .11 for the 7 that is the answer for part b what about for part c what is the probable that one random a select person brings between so the probable two and we know that the axis is the random variable for the population so the axis between 15 .6 than x and 16 .2.
02:05
Let me just graph the normal distribution here.
02:10
So the population has the distribution and the mean value for the population which is 16 and we have to get 15 .6 and 16 .2.
02:20
We have to get the area of this region.
02:23
So the probability of this region so we can find this value by using the normal cdf function of the calculator, normal cdf, lower boundary is 15 .6, upper boundary is 16 .2, and the mean value is 16, and the standard division, 6 .5.
02:43
Let's get the answer.
02:44
This is second distribution normal cdf, lower boundary, which is 15 .6, comma, this is 16 .2, 16, comma, and 6 .5.
02:59
We got the answer as 0 .368.
03:09
0368 that is the answer for part c and what about for part d for 34 people so we know that random variable for 34 people this is the sample standard division which is 16 and 1 .11 47 so we have to get the probability which is between 15 .6 and less than 16 .2 let's use the same function which is normal normal cdf here the lower boundary is 15 .6 the upper boundary is 16 .2 and the mean value is 16 and the standard division 1 .11 for the cell so we got the answer the second distribution normal cdf lower boundary this is 15 .6 comma this is 16 .2 comma and 16 comma this is 1 .11 which gives us the value of 0 .2113.
04:17
That is the answer for part d.
04:20
And what about for part d? the assumption is that the distribution is no more necessary.
04:27
Yes.
04:29
And what about the f? find the iker for the average of 34 coffee drinkers.
04:36
So we need to find the q1.
04:38
So the q1 means, which is 25th percentile, so that means the probability of x is less than x1, which has the probability of 0 .25.
04:52
In order to get the x1 value here, we need to get the z score.
04:57
So we have to use the inverse 9.
04:59
The area is 0 .25.
05:01
And the standard division for 34, i mean the sample size, which is actually for the standard z score, we have to use the standard normal distribution, which has the mean zero, and the standard division is 1.
05:17
So from here the z score, second distribution immersion norm.
05:22
0 .25 comma, this is 0 .1, which gives us the value of negative 0 .6745...