00:01
Now, in this question, you are giving that, you know, based on some historical data, your manager believes that 38 % of the company's orders come from first -time customers.
00:12
A random sample of 51 orders will be used to estimate proportion first -time customers, right? so what is the probability that the sample proportion is between 0 .2 and 0 .5? well, you know, if you look at the sample proportion, right, which i'm going to do by p, actually is likely to follow a normal distribution with the mean, of course, that's 0 .338, that's 38%.
00:34
And with a standard deviation, also the variance, that's actually going to be given by 0 .38 times 1 minus 0 .38, that's 0 .62, right? and then divided by 51.
00:48
So it's likely to follow this normal distribution, right? so if you do this calculation a bit, and you'll find this standard deviation is given by square roots, of 0 .38 times 6 .62 divided by 51.
01:04
And that actually gives you the sigma, which is standard deviation that's about 0 .068.
01:12
All right.
01:13
And from this, you're basically asked what's the probability, right? and with the p actually is between 0 .5 and 0 .2, right? now, you see that this 0 .5 actually is larger than the mean value.
01:29
While 0 .2 is actually less than the mean value.
01:32
So the probability can actually be conveniently represented by 1 minus alpha 1 minus al -2.
01:38
And alpha -1 -1 -2 you can actually obtain from the g scores at these two values.
01:44
So the z -square at alpha -1 is given by 0 .32 minus 0 .38 over the standard division, 0 .068...