00:01
They are given to invert investment here.
00:03
Let's say this is the investment one.
00:06
This is investment two.
00:08
And both of them has the normal distribution.
00:11
Let me just write the given information for the investments.
00:16
For the first one, the mean, so the mean, which is denoted by mo.
00:21
This is equal to $2 million.
00:23
So this is two times 10 to the bar of six.
00:27
And the standard deviation, which is standard.
00:30
Deviation which is denoted by sigma which is equal to 15 times so this is 15 times 10 to the power which is 4 and what about for the second one the second one is the mean this is 2 ,275 times 10 to the power which is 3 in the standard deviation which is 4 times 10 to the power 5 great.
01:08
And we can define the random variable x for this one, the normal distribution, mu and standard division.
01:13
And i can define the random variable y for the second investments, which is the mue and the standard division here.
01:20
Let's take a look at the first one here.
01:24
So the problem how the first investment will return less than this value.
01:31
So the x is less then which is 1 million this is 1 million 8 ,800 ,000 which is equal to so i'm going to use the normal cdf here the lower boundary which is negative 1 e99 the upper boundary 1 million 800 and the mean is 2 million and the standard division which is 15 1 2 3 4 great let's get the answer press second variance the second option here the lower boundary negative 1 this is 2nd e 99 and the upper boundary 18 so there are 2 4 5 zeros and this is 2 million okay this is 2 million great and the standard division which is 15 so we have 4 zeros so the answer would be which is this value here this is 0 .09 and 12 this is the answer we have for this question and the next one so for part b for the second investment, which is, so the probability of the second investment, which is 180, 800 ,000, which is again, i'm going to use the normal cdf.
03:00
So the lower boundary, this is 2, 275 and 3 zeros.
03:08
And the standard, so sorry, this is, so the upper boundary, sorry, the upper boundary is 1 ,800 ,000.
03:16
And the mean is 2 -275 -3 -0s and the standard division which is 400 ,000 let's get the answer just press second variance the normal cdf this is negative 1 second e99 and the upper boundary is 1 800 and this is 2 to 7530s and the 400 this one here so what about the probability the probability is 0 .11 and 75 great so we got this answer for the question and for part c we have to get the probability for both of them so by the way the second one should be y so the probability of random variable x which is less than this is 1 million 650 000 so this is 1 million 650 000 so i'm going to use the normal cdf and the lower boundary the upper boundary is 1650 and the mean is 2 million and the standard division 15.
04:34
Let's get the answer...