Question 1161766: Can you help me solve these. The instructor says I should be able to put it in my TI-89 calculator but I cannot get it to come out right.
1.) A manufacturer knows that their items have a normally distributed length, with a mean of 15.9 inches, and standard deviation of 1.2 inches.
If 17 items are chosen at random, what is the probability that their mean length is less than 15.8 inches?
2.) A manufacturer knows that their items have a normally distributed length, with a mean of 17.2 inches, and standard deviation of 2.6 inches.
If 4 items are chosen at random, what is the probability that their mean length is less than 17.2 inches?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! mean - 15.9 inches.
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.
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