SOLUTION: A trucking company determined that the distance traveled per truck per year is normally​ distributed, with a mean of 80 thousand miles and a standard deviation of 12 thousand mi
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Question 1191027: A trucking company determined that the distance traveled per truck per year is normally​ distributed, with a mean of 80 thousand miles and a standard deviation of 12 thousand miles.
How many miles will be traveled by at least ​70% of the​ trucks?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the mean is 80,000 and the standard deviation is 12,000.
the z-score for the probability that at least 70% of the trucks travel is equal to .5244.
the z-score formula is z = (x - m) / s
z = .5244
x = what you want to find.
m = mean = 80,000
s = standard deviation = 12,000
formula becomes:
.5244 = (x - 80,000) / 12,000.
solve for x to get:
x = .5244 * 12,000 + 80,000 = 86,292.8
replac4e x with that in the formula and you should get the z-score of .5244.
z = (86,292.8 - 80,000) / 12,000 = .5244.
find the area to the left of that z-score and you should get something very close to .7.
i got .6999998412.
you won't get .7 right on because .5244 is a rounded number.
the actual z-score that i got, using the ti-84 plus, was .5244005102.
in order to get a z-score of .5244 from the z-score tables, i had to do a manual interpolation.
the full z-score value from that was .5244092219.
i rounded to .5244.
the table only shows you the z-score rounded to 2 decimal digits.
what i got manually was actually better than that.
an online calculator i used, got me a z-score of .524.
if you rounded your z-score to what the table was s howing you, you would have gotten a z-score of .52.
that's because:
z-score of .52 gives you an area of .69847 to the left of it.
z-score of .53 gives you an area of .70194 to the left of it.
.70000 - .69847 = .00153.
.70194 - .70000 = .00194.
z-score of .52 gives you an area that is closer to .70000, so you would pick that.
here's the table i used.
https://www.rit.edu/academicsuccesscenter/sites/rit.edu.academicsuccesscenter/files/documents/math-handouts/Standard%20Normal%20Distribution%20Table.pdf
if you used the z-score of .52 from the table, you would have gotten a raw score of x = .52 * 12,000 + 80,000 = 86240.
i got 86,292.8 when rounding the z-score to 4 decimal digits.
the result are close, but different.
it depend on what your instructor expects when you solve for this.
if i used the calculator and didn't do any rounding, i would have gotten 86292.80612.
rounding to 4 decimal digits was definitely closer then rounding to 2.
let me know if you have any questions or concerns.
theo
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