SOLUTION: A population of values has a normal distribution with = 131.2 and o = 20.3. You intend to draw a random sample of size n = 18. Find the probability that a single randomly selecte

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Question 1204333: A population of values has a normal distribution with = 131.2 and o = 20.3. You intend to draw a random
sample of size n = 18.
Find the probability that a single randomly selected value is less than 142.7.
P(X < 142.7) =
Find the probability that a sample of size n = 18 is randomly selected with a mean less than 142.7.
P(M < 142.7) =
Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
population meaan is 131.2
population standard deviation is 20.3 (i believe that you meant standard deviation).
sample size is 18.

z-score formula is:
z = (x-m)/s
z is the z-score
x is the raw score
m is the raw mean
s is the standard deviation if you are looking for an individual score.
s is the standard error if you are looking for the mean of a sample of given size.

quesion 1.
Find the probability that a single randomly selected value is less than 142.7.

you use the z-score formula with the standard deviation to find the probability of finding a single score less than 142.7

z = (x-m) / s becomes z = (142.7 - 131.2) / 20.3.
you get z = .566502.
the probability of getting a z-score less than that is equal to .714474.
round to 4 decimal places to get .7145.

question 2.
Find the probability that a sample of size n = 18 is randomly selected with a mean less than 142.7.

use the z-score formula with the standard error to find the probability that the mean of a sample of size 18 is less than 142.7.

standard error = standard deviation / sqrt(sample size) = 20.3 / sqrt(18) = 4.784756.

z = (x - m) / s becomes:
z = (142.7 - 131.2) / 4.784756.
you get z = 2.403466
the probability of getting a z-score less than that is equal to .991880.
round to 4 decimal places to get .9919.

your solutions are:

the probability that a single randomly selected value is less than 142.7 is equal to .7145.

the probability that a sample of size n = 18 is randomly selected with a mean less than 142.7 is equal to .9919.

note that the standard error is defined as the standard deviation of the distribution of sample means.
that is not the same as the standard deviation of individual elements.
as the samaple size gets larger, the standard error becomes smaller.
that's bgecause the sample mean is closer to the population mean when the sample size is larger.

it can be dificult to visualize, but what you need to remember is:

if you are looking for the standard deviation of a single element in a set of data, then use the standard deviation.

if you are looking for the standard deviation of the mean of a sample of specified size from that set of data, then use the standard error.

note that, if the sample size is 1, then the standard error becomes the standard deviation because standard error = standard deviation / sqrt(1) = standard deviation / 1 = standard deviation.

i used a z-score calculator online to show you the difference between the z-sore of an individual element and the z-score of a sample mean of a sample of size 18.

the firest graph uses the standard deviation.
the second graph uses the standrd error.