00:01
It was found that 74 % of the aluminum cans sold in an area were recycled.
00:07
So that's a proportion of 0 .74, or a probability of 0 .74 for any randomly selected can being recycled.
00:15
We consider 381 cans that were sold today, and we're asked to use the normal approximation to find the probability that certain numbers of these cans will be recycled.
00:27
So if we define a random variable x as the number of cans out of the 381 that are recycled, then the number of cans that are recycled is a binomial random variable.
00:45
And when n times p is greater than 10, and n times 1 minus p is also greater than 10, which is the case here, then we can use the normal approximation to the binomial.
01:00
And the normal approximation is that the number is approximately normally distributed with a mean equal to n times p and a standard deviation equal to the square root of n times p times 1 minus p.
01:23
So if we calculate these, the approximation is that x is normally distributed with a mean of 281 .94 and a standard deviation of 8 .5618 approximately.
01:40
For part a we want the probability that 300 or more cans will be recycled.
01:49
So this is the probability that x is greater than or equal to 300.
01:53
First let's express this in terms of a cumulative probability.
01:57
Using the complement rule this is 1 minus the probability that x is at most 299.
02:06
Then since we're using a continuous distribution to model a discrete distribution we can apply a continuity correction factor and express this as 1 minus the probability that x is at most 299 .5...