00:01
It's given in this exercise that the weights of a certain laboratory population of mice at 20 days, let's call this random variable x, are approximately normally distributed with a mean of 8 .3 grams and a standard deviation of 1 .7 grams.
00:19
We consider samples of size 10 taken from this population, and we consider the total weight of the samples, or the total weights of the samples.
00:29
We are asked for the probability that the total weight of a sample is at least 80 grams.
00:37
So t is a random variable because it's the sum of 10 random variables which have these parameters.
00:47
So we can say t is equal to the summation of 10 of these random variables, which are the heights which are the weights of the mice.
01:08
So in order to answer this question, we have to know how t is distributed.
01:17
Again, t is the sum of 10 independent and identical random variables, 10 x's.
01:27
The mean of t or the expected value of t is the expected value of this expression.
01:45
And this is equal to, it's equal to this, and that is equal to 10 times the expected value of an individual mice, individual mouse.
02:14
And so this is 10 times 8 .3 or 83 grams.
02:21
So the mean of the distribution of total weights of 10 mice is 83 grams.
02:31
And now all we need now is the standard deviation of t.
02:53
When we are summing random variables, their variances add, or their standard deviations combine in this method.
03:03
Their standard deviations combine in this manner.
03:28
Now, since the standard deviation is 1 .7, the variance for each individual mouse randomly selected is 1 .7 squared.
03:46
So it's 10 times 1 .7 squared...