SOLUTION: Suppose the length of a species of crocodiles is normally distributed. If 99.7% of these crocodiles are between 7.5 and 20.7 feet long, determine the standard deviation of the dist

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Question 1102155: Suppose the length of a species of crocodiles is normally distributed. If 99.7% of these crocodiles are between 7.5 and 20.7 feet long, determine the standard deviation of the distribution in feet.
Answer by Theo(13342) About Me  (Show Source):
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99.7% of the crocodiles are between 7.5 and 20.7 feet.

their length is normally distributed.

since the normal distribution is symmetric about the mean, then the mean must be equal to 20.7 minus (20.7 - 7.5) / 2 = 20.7 minus 6.6 = 14.1

since .997 of the area under the normal distribution curve is between 7.5 and 20.7, then .003 of the area under the normal distribution curve must be outside of these limits.

take half of that to get .0015 of the area under the normal distribution curve is to the right of the upper limit and .0015 of the area under the normal distribution curve is to the left of the lower limit.

to find the z-score associated with this, find the area to the left of the upper limit.

the area to the left of the upper limit is 1 - .0015 = .9985.

use a z-score calculator to find that the z-score associated with this is z = 2.967737927

since the normal distribution is symmetric about the mean, then the lower limit z-score will be -2.967737927

since 20.7 is 6.6 feet more than the mean of 14.1, then 6.6 feet must be 2.967737927 standard deviations above the mean.

solve for the standard deviation to get standard deviation = 6.6 / 2.967737927 = 2.223916047 feet.

the mean length of the crocodiles is 14.1 feet and the standard deviation is 2.223916047 feet.

this can be seen to be true from the following normal distribution calculator printout.

$$$

the mean is 14.1
the standard deviation is 2.223916 which is the same as 2.223916047 except it is rounded to 6 decimal digits.
the calculator is requested to find the area between a lower limit of 7.5 and an upper limit of 20.7
the calculator says that the area under the normal distribution curve between those two limits is .997 which is the same as 99.7%.