00:01
So in this question we're told that a, that there is a 0 .15 chance of a transcription error, transmission error, for each bit transmitted.
00:25
And in part a, we're told that there are n equals 1 ,000 bits transmitted.
00:30
So that means the number of transcription errors is going to be binomially distributed with n equals 1 ,000 and p equals 0 .15.
00:40
But for numbers as big as this, what we can say is that np is going to be 150, which is the mean.
00:50
And the variance is np 1 minus p, which is 150 times 0 .85, which is 127 .5.
01:02
These are both large.
01:05
So we can approximate n as being normally distributed with mean 150 and variance, 127.
01:14
So this is an approximate distribution.
01:22
So what's the probability that at most 165 transmission errors occur? well, let's construct a variable z, which is n minus mu over sigma, then this is going to be normally distributed with mean 0 and variance 1.
01:48
Then the probability that n is less than or equal to 165 is going to be equal to the point of the probability that probability that z is less than or equal to 165 minus 150 divided by the square root of 127 .5, which is 1 .328 standard deviations above the mean.
02:14
So we're going to go to a z score calculator and plug in the probability that z is less than 1 .328.
02:23
And this is going to give us a probability of 0 .9079.
02:30
So we've got a 90 .79 % chance of there being less than a equal to 165 transcription errors...