00:01
So, dense as per the given question x is a normal random variable with mean mu is equals to 10 and variance sigma square is equals to 4.
00:12
So, we need to find probability of x greater than 10.
00:18
So, we need to calculate the standard deviation which is sigma and the square root of variance is equals to sigma is equals to root 4 is equals to 2.
00:27
So, now we have standard deviation as well.
00:29
So, let us substitute all these values in z formula.
00:32
Standard normal distribution z is given by z is equals to x minus mu by sigma is the standard formula which is equals to 10 minus 10 by 2 which is equals to 0.
00:46
So, the probability that p of x is greater than 10 is equivalent to area of right of z is equals to 0 which is 0 .5.
01:01
So, in bit b we need to find the probability of x is less than 12.
01:06
So, we use the same standard normal distribution probability formula where z is equals to 12 minus 10 by 2 which is equals to 1.
01:17
So, the probability of p of x is less than or equals to 12 is the area which is left of z is equals to 1 which is approximately equals to 0 .8413 and for this it is right z is equals to 0 and which is 0 .5.
01:39
And for c we need to find the probability of p of 10 less than x less than 12.
01:48
So, we can subtract the probability of x is less than 12 and the probability of x is less than 10...