00:01
So, for this problem, to begin, i'll note that we'll be working from the fundamental idea that the sample means, when the population is normally distributed, the sample means will follow a normal distribution with a mean value equal to that of the population, in this case 4064, and will have a standard deviation equal to that of the population, 460, oh, not t60, 460, divided by the square root of the sample size.
00:30
Both of the problems are with sample size 20, so we'd have a standard deviation of sample means of roughly 102 .9.
00:41
So, for part a, define probability of x bar less than 3750.
00:48
Can use our ti -84 here, so we hit 2nd, vars or distr, normal cumulative distribution function, or normal cdf, lower, yeah, leave it as arbitrarily large negative number, upper, 3750, mean value 4064, oops, 4064, sigma, 102 .9, paste, hit enter, and we get a result of 0 .0011.
01:16
Then, for part b, define the probability of x bar greater than 4125...