SOLUTION: As reported in "Runner's World" magazine, the times of the finishers in the New York City 10 km run are normally distributed with a mean of 61 minutes and a standard deviation of 9
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Question 1204356: As reported in "Runner's World" magazine, the times of the finishers in the New York City 10 km run are normally distributed with a mean of 61 minutes and a standard deviation of 9 minutes. Let x denote finishing time for the finishers.
a) The distribution of the variable x has mean
and standard deviation
.
b) The distribution of the standardized variable z has mean
and standard deviation
.
c) The percentage of finishers with times between 40 and 65 minutes is equal to the area under the standard normal curve between
and
.
d) The percentage of finishers with times exceeding 86 minutes is equal to the area under the standard normal curve that lies to the
of
. Answer by Theo(13342) (Show Source):
excerpt from the web:
A standardized variable (sometimes called a z-score or a standard score) is a variable that has been rescaled to have a mean of zero and a standard deviation of one.
formula for z-score is z = (x-m)/s
x is the variable.
m is the mean
s is the standard deviation.
z is the standardized variable, called the z-score.
the standardization allows different sets of data with different means and standard deviations to be compared to each other.
the z-score tells you how many standard deviations you are from the mean.
the mean of the variable is 61 and the standard deviation is 9.
the mean of the standardized variable is 0 and the standard deviation is 1.
z-score formula is z = (x-m)/s
z is the z-score
x is the variable score
m is the mean
s is the standard deviation.
when x is 40, z = (40 - 61) / 9 = -2.3333.
when x is 65, z = (65 - 61) / 9 = .44444.
the percentage of finishers with times between 40 and 65 minutes is equal to the area under the normal distribution curve between z-scores of -2.3333 and .4444.
the z-scores are usualy rounded to between 2 and 4 decimal digits.
there are calculators to help you find the area in bgetween directly.
otherwisee you would find the area to the left of the low z-score and to the left of the high z-score and subtract the smaller area from the larger area to get the area in between.
to find the percentages of finishers that exceed 86 minutes, find the z-score for x = 86 and then find the area to the right of that.
z = (86 - 61) / 9 = 2.7778.
find the area to the right of that.
i rounded all z-score to 4 decimal digits which provides sufficient accuracy for most situations.
this calculator allows you to find values from the raw scores and from the z-scores.
i'll show you the results from it that were obtained from the z-scores amd from the raw scpres.
they'll either be the same or close to each other, depending on rounding differences.
first is finding the area in between 40 and 65 using z-score and then raw scores.
next is finding the area to the right of 86 using z-scores and then raw scores.
in the calculator, when you're working with z-score, you set the mean to 0 and the standard deviation to 1; when you're working with raw scores, you set the mean and standard deviatiion to whatever they are for the problem you are working.
answers to your questions.
a) The distribution of the variable x has mean of 61 and standard deviation of 9.
b) The distribution of the standardized variable z has meanof 0 and standard deviation of 1.
c) The percentage of finishers with times between 40 and 65 minutes is equal to the area under the standard normal curve between z = -2.3333 and .4444.
d) The percentage of finishers with times exceeding 86 minutes is equal to the area under the standard normal curve that lies to the right of z-score of 2.7778.
