00:01
A normal distribution is given here.
00:03
The mean value, which is denoted by mu, which is given as 42 .8 pounds.
00:12
And the standard deviation, which is denoted by sigma, that was given as 2 .8 pounds.
00:19
Let me define the random variable x here, which is normally distributed.
00:24
So it has the mean value for the 2 .8 and the standard division 2 .8.
00:29
So we got these results here.
00:31
So for the first one, we have to get the probability of a dog that weights more than 45 pounds.
00:39
So for part a, i need to find the random variable x, which is greater than this is 45 pounds.
00:46
So what am i supposed to do? let me just graph the normal distribution first, and i can just easily understand where i need to find.
00:53
Here is the mean score, which is 42 .8, and the 45 is.
00:59
Here so we have to get this shaded region so if you have the graphing this lay calculator we can easily find by using the normal cdf function what we have to put we have to put the lower boundary for the shaded region that is 45 and for the upper boundary that goes to infinity so i'm going to put 1a99 and the mean value is 42 .8 and the standard division which is 2 .8 let's get the answer just press second and distribution there is normal cdf here lower boundary 45 the upper boundary is one second in 99 and the mean value is 42 .8 and the standard division which is 2 .8 let's get the answer this is 0 .21 and 660 that is the answer that we have for part a and what about for part b it says find the value that corresponds to the 18th percentile of the distribution.
02:00
So what that means? the 18th percentile means, let's say this is equal to x1.
02:08
So that means if you just graph the normal distribution, there should be some x1 here.
02:14
So the whole region, shaded region, should be the 80 percentile, which is 0 .8.
02:21
So we need to get the value of x1 here.
02:23
So the probability of x is less than x1 which is 0 .8 so in order to find the x1 value here i need to find the standard z score so i'm going to use the inverse norm function here the area is 0 .8 0 .8 and the mean value is 0 standard division is 1 so from here the z value would be so i'm going to use the second distribution inverse norm the area is 0 .8 and the mean value is standard division is 1.
02:52
That would be 0 .84 and 16.
02:56
To get the x1 value, i'm going to use this formula, which is x minus mu over standard division.
03:02
This is the formula.
03:03
I'm going to just plug in all the values to this formula.
03:05
This is 0 .84, 16, which is equal to x1 minus.
03:10
The mean value was 42 .8 and the standard division 2 .8 here, just to cross multiplication and leave x1 and 1.
03:18
I'm going to do this calculation by calculator.
03:21
Just multiply with 2 .8 and plus 42 .8...