Refer to the Buena School District bus data a. Refer to the maintenance cost var
ID: 3180193 • Letter: R
Question
Refer to the Buena School District bus data
a. Refer to the maintenance cost variable. The mean maintenance cost for last year is $450.29, with a standard deviation of $53.69. Estimate the number of buses with a cost of more than $500. Compare that with the actual number.
b. Refer to the variable on the number of miles driven. The mean is 830.11 and the standard deviation is 42.19 miles. Estimate the number of buses traveling more than 900 miles. Comparethat number with the actual value.
Frequency Distribution - Quantitative Maintenance cumulative lower upper midpoint width frequency percent frequency percent 320 < 340 330 20 3 3.8 3 3.8 340 < 360 350 20 3 3.8 6 7.5 360 < 380 370 20 1 1.3 7 8.8 380 < 400 390 20 8 10.0 15 18.8 400 < 420 410 20 5 6.3 20 25.0 420 < 440 430 20 12 15.0 32 40.0 440 < 460 450 20 12 15.0 44 55.0 460 < 480 470 20 15 18.8 59 73.8 480 < 500 490 20 7 8.8 66 82.5 500 < 520 510 20 8 10.0 74 92.5 520 < 540 530 20 1 1.3 75 93.8 540 < 560 550 20 3 3.8 78 97.5 560 < 580 570 20 2 2.5 80 100.0 80 100.0 Descriptive statistics Maintenance count 80 mean 450.29 sample variance 2,882.18 sample standard deviation 53.69 minimum 329 maximum 570 range 241 1st quartile 420.00 median 456.00 3rd quartile 489.75 interquartile range 69.75 mode 329.00 low extremes 0 low outliers 0 high outliers 0 high extremes 0 Frequency Distribution - Quantitative Maintenance cumulative lower upper midpoint width frequency percent frequency percent 300 < 350 325 50 3 3.8 3 3.8 350 < 400 375 50 12 15.0 15 18.8 400 < 450 425 50 22 27.5 37 46.3 450 < 500 475 50 29 36.3 66 82.5 500 < 550 525 50 11 13.8 77 96.3 550 < 600 575 50 3 3.8 80 100.0 80 100.0 Frequency Distribution - Quantitative Maintenance cumulative lower upper midpoint width frequency percent frequency percent 325 < 360 343 35 6 7.5 6 7.5 360 < 395 378 35 8 10.0 14 17.5 395 < 430 413 35 12 15.0 26 32.5 430 < 465 448 35 20 25.0 46 57.5 465 < 500 483 35 20 25.0 66 82.5 500 < 535 518 35 9 11.3 75 93.8 535 < 570 553 35 4 5.0 79 98.8 570 < 605 587 35 1 1.3 80 100.0 80 100.0 Descriptive statistics Miles count 80 mean 827.53 sample variance 1,188.20 sample standard deviation 34.47 minimum 741 maximum 908 range 167 1st quartile 806.00 median 827.00 3rd quartile 851.50 interquartile range 45.50 mode 815.00 low extremes 0 low outliers 0 high outliers 0 high extremes 0 1/28/2010 16:03.39 (1) Descriptive statistics Miles count 80 mean 827.53 sample variance 1,188.20 sample standard deviation 34.47 minimum 741 maximum 908 range 167 1st quartile 806.00 median 827.00 3rd quartile 851.50 interquartile range 45.50 mode 815.00 low extremes 0 low outliers 0 high outliers 0 high extremes 0 1/28/2010 16:06.08 (1) 1/28/2010 16:06.08 (1) Descriptive statistics Miles count 80 mean 830.11 sample variance 1,779.85 sample standard deviation 42.19 minimum 741 maximum 1008 range 267 1st quartile 806.00 median 827.00 3rd quartile 851.50 interquartile range 45.50 mode 815.00 low extremes 0 low outliers 0 high outliers 1 high extremes 1 1/28/2010 16:08.20 (1) 1/28/2010 16:08.20 (1) Correlation Matrix Maintenance Age Miles Maintenance 1.000 Age .465 1.000 Miles .450 .522 1.000 80 sample size ± .220 critical value .05 (two-tail) ± .286 critical value .01 (two-tail)Explanation / Answer
a. Refer to the maintenance cost variable. The mean maintenance cost for last year is $450.29, with a standard deviation of $53.69. Estimate the number of buses with a cost of more than $500. Compare that with the actual number.
we are given that mean = 450.29 and sd = 53.69 and we need to find P(X>500) , once we have the probability we can multiply it by the number of buses to arrive at the answer so we first calculate the z score
using the formula
Z = (X - Mean)/SD
(500-450.29)/53.69 = 0.925 so using z tables we need to find P(Z>0.925)
P ( Z>0.925 )=1P ( Z<0.925 )=10.8238=0.1762
now if the total number of buses is 80 , so the number is 80*0.1762 = 14 buses approx
please keep the z tables handy for the above calculations
b. Refer to the variable on the number of miles driven. The mean is 830.11 and the standard deviation is 42.19 miles. Estimate the number of buses traveling more than 900 miles. Comparethat number with the actual value.
again using the same concept as above
Z = (900-830.11)/42.19 = 1.656
P(Z>1.656)
we get the value as
P ( Z>1.656 )=1P ( Z<1.656 )=10.9515=0.0485
number of buses = 80
so number of buses with miles greater than 900 are 0.0485*80 = 4 approx
The data is not given properly , so i have assumed the number of buses to be 80 here