00:01
There is given a binomial distribution in this question.
00:02
So the n, the sample size, 190, and the probability here which is 0 .78.
00:07
Because the sample size is very large, so we have to convert this binomial distribution into the normal distribution.
00:13
So what i need? i need the mean value which is n times p, so this is 190 times 0 .78, and multiply by 0 .78, which would be 148 .2.
00:28
And also i need to get the standard deviation.
00:30
For the binomial distribution, the standard deviation is n times p times 1 minus p.
00:35
So this is 190 times 0 .78 n times 1 minus 0 .78, which is 0 .22.
00:42
So this is the square root of 190 n times 0 .78 n times 0 .22.
00:50
So the value here which is 5 .71.
00:52
So we can define a random variable x bar, which is normally distributed, the mean, and the standard deviation 5 .71.
00:59
Let's calculate each of them.
01:02
So probability of getting the x bar, which is less than or equal to 130.
01:06
But be careful, we converted the binomial distribution into a normal distribution.
01:11
So we have to apply the continuity correction at this step.
01:15
So if i add 0 .5 here, so we have to find the area which is less than 130 .5.
01:21
To get this probability, i'm going to use a normal cdf function.
01:24
There is no lower boundary, but for a small number, upper boundary, the mean and the standard deviation for the standard normal distribution.
01:32
So, sorry, the mean here, the mean for this distribution, 148 .2, and the standard division, 5 .71.
01:41
Press second, variance, the normal cdf, lower boundary, this is negative 1, second, 899.
01:46
The upper boundary is 130 .5 and the mean is 148 .2, and the standard division, which is 5 .71.
01:53
So the probability would be, which is 0 .00, this is 10 .0 with four decimal places...