00:01
A normal distribution question is given here.
00:03
Let me just take some notes.
00:05
The mean weight here, let me just write as mean, which is denoted by the sample of mu here, that is 5 .67.
00:15
And the standard deviation, which is denoted by sigma, that is 0 .0602.
00:24
So in the first part, let's say since this is a normal distribution, i can define the x as the random variable, normally distributed, that is 5 .67 and 0 .062.
00:37
So in the first part, we have to get the probability, which is between 5 .55, less than x, less than, and this is 5 .7 .5 .59.
00:51
We need to get this probability here.
00:53
Let me just graph the normal distribution, and we can easily understand what that means.
00:58
So this is 5 .67.
01:00
So here, is 5 .55 and here is 5 .7 and 9 so we need to get the area of this shaded region.
01:10
How can i get? i'm going to use i'm going to use the function for the ti calculator which is normal cdf.
01:21
So i'm going to use the normal cdf function of the calculator.
01:26
So i have to put the lower boundary here which is 5 .55.
01:31
The upper boundary is 5 .79 and the mean value is 5 .67 and the standard deviation 0 .062.
01:40
Let's get the answer.
01:42
For normal cdf, just press the button second end distribution.
01:45
There's normal cdf here.
01:46
Lower boundary is 5 .55 comma, the upper boundary is 5 .79, comma, and 5 .67 and comma 0 .062.
01:59
We got the probability as 0.
02:02
This is 9471.
02:06
That is the answer for part a.
02:09
What about for part b? so in part b, if 4 randomly selected.
02:15
So the n in this case is given as 4.
02:18
So the sample mean score, let me just use x bar.
02:24
This is the sample mean, which is equal to the population mean here...