00:01
In this question, there is given a binomial distribution, or the binomial probability distribution here.
00:07
So the sample size, which is n, that was given as 900, and the probability of success, which is p, denoted by p, this is equal to 0 .50.
00:16
So we can get the value of q here, so the q is equal to 1 minus p.
00:21
That means this is 1 minus 0 .50, which is also 0 .50.
00:25
So we got the q value.
00:27
So we're going to use the standard normal distribution, i mean normal distribution, for getting the probabilities here.
00:36
So what we need? we need the mean score.
00:38
So the mean, which is equal to, this is n times p.
00:42
So the mean, which is denoted by mu, this is n is 900 times 1 over 2 or 0 .50, which is 1 over 2, this is equal to 450.
00:54
And what about for the standard deviation? the binomial distribution, standard deviation is equal to n times, p times q.
01:03
So the standard deviation denoted by sigma, this is equal to 900 times 0 .5 times again 0 .5, which is given, this is square root of 900 times 0 .5 times 3 .5.
01:20
So it is equal to 15.
01:22
So i can define the random variable x here, which is normally distributed.
01:25
So the mean value was, first, 450 and the standard division which is 15.
01:31
So what we have to find in part a, in part a, we have to find the probability of the probability of axis greater than 490.
01:42
So i'm going to use the normal cdf function to get this probability.
01:46
The lower boundary here, 490.
01:48
There is no in upper boundary.
01:49
I'm going to put a big number.
01:51
The mean value we got as 450 and the standard division is 15.
01:56
Let's get the answer.
01:56
This is second and variance.
01:59
The second option i'm going to select.
02:01
This is 490.
02:03
This is one second e99.
02:06
And this is 450 and we got 15.
02:10
So the answer would be 0 points.
02:13
So how many decimal points do we need? four decimal places.
02:16
This is 0 .00 and 38.
02:21
What about for part b? so the probability, which is fever then, that means the probability of random variable x this is less than 425 again i'm going to use the normal cdf function here there is no any lower boundary here so i'm going to put very small number the upper boundary is 425 the mean and the standard division let's get the answer this is second variance normal cdf negative 1 this is second a 99 and the upper boundary is 425 this is 450 and 15 so so the answer would be 0 .04 and 78.
03:02
And what about for part c? this is an interval, so the probability of random variable x.
03:08
This is between 445 and it is 480.
03:15
Again let me use the normal cdf here, the lower boundary, 445, this is 480, the mean value, and the standard division.
03:27
Let's get the answer again.
03:28
This is second variance, the normal cdf, 445, this is 480, and then 450, and then 15.
03:37
So the answer is 0.
03:40
This is 0 .0.
03:44
0 .60 and 78...