standard deviation is 1.2 inches.
sample is 17 items.
standard deviation of distribution of sample means (standard error) is equal to standard deviation / sqrt (sample size) = 1.2 / sqrt(17) = .29104275.
i have a ti-84 which is similar to ti-89
you can find it by z-score or you can find it by raw score.
to find it by z-score, then calculate z-score.
formula is z = (x-m)/s = (15.8-15.9)/.29104275 = -.3435921355
find area to the left of that.
command is normalcdf(-E9,-.3435921355) = .365576582.
that's your answer.
alternatively, you can enter the raw score and the mean and the standard error directly as normalcdf(-E9,15.8,15.9,.29104275) = .365576582.
first method used z-score.
second method used raw score.
-E9 means -10^9 which is the area to the left of the z-score (first method), or area to the left of the raw score (second method).
if you did everything else correctly, i suspect you didn't get the standard error correctly, and possible used the standard deviation instead, but that's only conjecture.
for your second problem:
mean is 17.2 and standard deviation is 2.6.
4 items are chosen at random, therefore sample size is 4.
probability of mean length should be .5 since,in a normal distribution, the mean is at the midpoint of the distribution.
still, ........
standard error = standard deviation divided by sqrt of sample size = 2.6 / sqrt(4) = 1.3
use the second method to make entries as normalcdf(-E9,17.2,17.2,1.3) = .5
in this particular case, since you were at the mean, the standard error didn't matter because the numerator in the z-score equation came out as 0.
for example:
z = (17.2 - 17.2) / se = 0 / se = 0.
se stands for standard error which is also called the standard deviation of the distribution of samplemeans.
the z-score of 0 is the same as the mean.
the mean in a normal distribution is always at the midpoint of the distribution curve with half of the area to the left and half of the area to the right.
i also used an online calculator to confirm the results.
here are the displays from that calculator.
4 displays.
2 from first problem and 2 from second problem.
first of each set of 2 uses z-score.
second of each set of 2 uses raw score.
in both cases, the standard error is calculated as standard deviation / square root of sample size.
these answers should be correct if i understood the problem correctly.
let me know is this satisfies your inquiry.