00:01
So in this question we're told that for a given customer, the probability of taking the sample is 58%.
00:13
And for all those who take the free sample, the probability of buying, given that you take the sample, is 41%.
00:24
Just for consistency in my notation, i'll say this is probability of sampling.
00:29
So on a given day, we have n equals 309 customers.
00:41
So first of all, what's the probability that more than 180 will take a free sample? so the numbers who take the sample, ns, is going to be binomially distributed with 309 trials and a 58 % chance of success.
01:00
And can we approximate this with the normal? well, the mean is 309 times 0 .58, which is a 109.
01:09
179 .22.
01:11
And the variance is the mean multiplied by the chance of failure, 0 .6, sorry, 0 .42, which is 75 .2, sorry, 2724.
01:28
So these are both large enough.
01:30
So both much larger than, say, 5 or 10.
01:37
So the normal approximation is appropriate.
01:48
So that means that the number that sample is approximately normally distributed with mean 179 .22 and variance 75 .2724.
02:03
So the probability, so first of all, let's say that z sample is the number of samples minus the mean divided by the standard deviation.
02:13
And the probability that the number of people that sample is more than 180 is equal to the probability that the z of samples is more than 180 minus 179 .22 divided by 75 .2724, which is 0 .0104.
02:41
And if we go to a z score calculator and we put in a z score of 0 .0104, then we get that the probability that x is greater than that is 0 .49585.
03:01
And since we're meant to round to four decimal places, i believe, yes, then that gives us 0 .499 as the approximate probability.
03:16
What's the probability that fewer than 200 will take the free sample? probability that ns is less than 200 is the probability that the z score of the samples is less than 200 minus 179 .22 divided by, sorry, the square root of 75 .2724, which is 2 .39 .951...