Question

A trucking company determined that the distance traveled per truck per year is normally distributed, with a mean of 30 thousand miles and a standard deviation of 11 thousand miles. Complete parts (a) through (d) below. a. What proportion of trucks can be expected to travel between 17 and 30 thousand miles in a year? The proportion of trucks that can be expected to travel between 17 and 30 thousand miles in a year is (Round to four decimal places as needed.) b. What percentage of trucks can be expected to travel either less than 15 or more than 40 thousand miles in a year? The percentage of trucks that can be expected to travel either less than 15 or more than 40 thousand miles in a year is % (Round to two decimal places as needed.) c. How many miles will be traveled by at least 85% of the trucks? The number of miles that will be traveled by at least 85% of the trucks is miles. (Round to the nearest mile as needed.) d. What are your answers to parts (a) through (c) if the standard deviation is 8 thousand miles? If the standard deviation is 8 thousand miles, the proportion of trucks that can be expected to travel between 17 and 30 thousand miles in a year 0 (Round to four decimal places as needed.) If the standard deviation is 8 thousand miles, the percentage of trucks that can be expected to travel either less than 15 or more than 40 thousand miles in a year is %. (Round to two decimal places as needed.) of the standard deviation is 8 thousand miles, the number of miles that will be traveled by at least 85% of the trucks is miles. ind in the nearest mile se noorart

Answer 1

Answer: Given = 30 11 o mean (M) standard dev (o)= The proportion of toucky can be expected to travel between 17 and 30 thousand miles in a year is 30-30 P(17 2 * 2-30) 17-30 = X-M L [ ** 11 :P (-1.1824 Z 20) = P(220) - PLZ 2-1.182) : 0.5 -0.1186 (from 2-table) = 0.3814 Less than X-M propor tion = 0.3814 The percentage of trucks can be expected to travel either is 15 (87) more than 40 thousand miles in year P (x215) =p 15.739 Plze -1.364) 0.0863 (from -table) ap(x215) = 8.63% 40-30 PCX > 40) =p p[X-4 2- 11 P(Z > 0.91) = 1- P(Z 20.91) S1-0.8186 (From Z- -table) - O: 1814 ... P(x>40) = 18.14%,

The number of milles that will be traveled by at least 85% of the trucks as P(Z >2) 854. 1- P(Z = z) = 0.85 p(222) = 1-0.85 P(Z <Z) = 0.15 P (24-104) = 0.15 From z-table Z = -1.04. Now using z- score formula Х = (Z *) + M =(-1004 * 11) + 30 - - 11.44 + 30 18:56 х = 18 miles. If the standard deviation (0) = 8 X-M L 17-30 30-30 8 (i) P(87 17.30 X LX 230) - P P(-1.625 2 240) = P (220) - P(Z2-1635) os 0.0521 ::P = 0.4479. P(172x230) = 0.4479 (ii) P(x215) = p[zclado = P(ZL -1.875) 0.0304 P/X215) = 3.04%

40-30 8 PLX 340) = P[2> P(Z > 1.25) P(Z 21:25) (From 2- Lable) = =1-0.8944 11 o.1056 i p(x>40) = 10:56 y. P(Z >2) = 85%. 0.85 - (242) d P(Z <Z) = 0.15 P(Z 2-1.04) -0.15 Z = -1.04 Formula x = (Z*-) +M (-1.04 *8) + 30 - -8.32 + 30 = 21.68 x = 21 miles.