population mean = 205.5
population standard deviation = 8.6
the first part of your question deals with a single individual and the probability that the individual will have a height between 179.7 cm and 231.3 cm.
since this is an individual score and not the mean of a sample, you use the standard deviation of the population.
first you find the z-score and then you find the area under the curve.
the z-score is found by the following formula:
z = (x-m)/s
x is equal to the actual score.
m is equal to the mean of the population.
since you are looking for the area of the normal distribution curve between two limits, those two limits becomes the x in this formula.
you will have a z-score using the lowest height limit as x, and you will have a z-score using the highest height limit as x.
the z-score for the low height is equal to (179.7 - 205.5) / 8.6 which is equal to -25.8 / 8.6 which is equal to -3.
the z-score for the high height is equal to (231.3 - 205.5) / 8.6 which is equal to 25.8 / 8.6 which is equal to 3.
you look up the z-score table to find the area under the normal distribution curve that is to the left of those z-scores.
some of the tables work differently than others, so make sure to follow the instructions of the table.
the table i am using is at the following link:
http://lilt.ilstu.edu/dasacke/eco148/ztable.htm
this table gives you the area under the curve to the left of the z-score.
the z-score of -3 gives you an area under the normal distribution curve to the left of that z-score as .0013
the z-score of 3 gives you an area under the normal distribution curve to the left of that z-score as .9987
you subtract the lower z-score area value from the higher z-score area value to get the area between them.
that area is .9987 - .0013 = .9974.
that's your answer for the first part of your question.
the probability of her height being between the low and high height is .9974.
the second part of your question deals with a sample of 40 and the probability that the sample mean is between 204 cm and 206 cm.
because you are dealing with a distribution of sample means, you will need to use the standard error formula to get the standard error of the distribution of means and then pretty much do the same thing as you did for the individual.
the only difference is the standard error is not the same as the standard deviation in this case, and you are dealing with the mean of the sample rather than the score of an individual.
the formula for standard error is:
se = sd / sqrt(n)
sd = 8.6
n = 40
se = 8.6 / sqrt(40) which is equal to 1.3598
you are looking for a low height of 204 and a high height of 206.
your standard error is 1.3598.
your population mean is still 205.5
the z-score is found by the following formula:
z = (x-m)/s
x is equal to the mean of the sample.
since you are looking for the area of the normal distribution curve between two sample mean limits, those two limits becomes the x in this formula.
you will have a z-score using the lowest sample height limit as x, and you will have a z-score using the highest sample height limit as x.
the z-score for the low height is equal to (204 - 205.5) / 1.3598 which is equal to -1.5 / 1.3598 which is equal to -1.1031.
the z-score for the high height is equal to (206 - 205.5) / 1.3598 which is equal to .5 / 1.3598 which is equal to .3677.
you look up the z-score table to find the area under the normal distribution curve that is to the left of those z-scores.
that z-score table, can be found, once again, at the following link:
http://lilt.ilstu.edu/dasacke/eco148/ztable.htm
since the tables are only accurate to 2 decimal places, round your z-score to 2 decimal places before using the table, if it is more detailed than that.
-1.1031 was rounded to -1.10
.3677 was rounded to .37
the z-score of -1.10 gives you an area under the normal distribution curve to the left of that z-score as .1357.
the z-score of .37 gives you an area under the normal distribution curve to the left of that z-score as .6443.
you subtract the lower z-score area value from the higher z-score area value to get the area between them.
that area is .6443 - .1357 which is equal to .5086.
if you want to see what these look like under the normal distribution curve, the following pictures may be helpful.
the numbers may not be exact, but any difference is due to rounding.
