Question

using normal distribution with steps please

SELF (6.8 Toby's Trucking Company determined that the dis- TEST tance traveled per truck per year is normally distributed, with a mean of 50 thousand miles and a standard deviation of 12 thousand miles. a. What proportion of trucks can be expected to travel between 34 and 50 thousand miles in a year? b. What percentage of trucks can be expected to travel either less than 30 or more than 60 thousand miles in a year? c.How many miles will be traveled by at least 80 percent of the trucks? d. What are your answers to (a) through (c) if the standard deviation is 10 thousand miles?

Answer 1

Solution: Мean %3D и 3 50 Standard deviation a= 12 Part-A: Proportion of trucks that can be expected to travel between 34 and 50 thousand miles P(34 x 50) Z-score for (x = 34): = 2 34 50 Z = 12 z =-1.333 Z-score for (x = 50): 1- x Z= 50 50 Z = 12 z= 0 Hence, P(34 x < 50) P(-1.333 z 0) P(z < 0) P(z -1.333) From normal distribution table; P(z 0) - P(z -1.333) = 0.5000 0.0912 = 0.4088 Therefore, proportion of trucks that can be expected to travel between 34 and 50 thousand miles 0.4088 Part-B: Proportion of trucks that can be expected to travel either less than 30 or more than 60 thousand miles = Р(х < 30) + Р(х > 60) Z-score for (x = 30): = Z 30 50 Z = 12 z =-1.667

Z-score for (x = 60): Z= 60 50 12 Z=0,833 Hence, P(x < 30)P(x> 60) P(z <-1.667)P(z> 0.833) From normal distribution table; P(z < -1.667) P(z > 0.833) = 0.0478 +0.2023 0.2501 Therefore, proportion of trucks that can be expected to travel either less than 30 or more than 60 thousand miles = 0.2501 Part-C: Value of z-score, such that area under the normal curve to the right of it equal to 80%, is given by: z = -0.842 Z = x-50 -0.842 = 12 x= 39.9 Hence, number of miles travelled by at least 80 percent of the trucks 39.9 thousand Part-D: g 10 Proportion of trucks that can be expected to travel between 34 and 50 thousand miles = P(34 < x < 50) Z-score for (x = 34): 34 50 Z = 10 z = -1.600

Z-score for (x = 50): = Z 50 50 Z= 10 z= 0 Hence, P(34 < x < 50) P(-1.600 z < 0) P(z 0) - P(z -1.600) From normal distribution table; P(z 0) - P(z <-1.600) 0.5000 0.0548 0.4452 Therefore, proportion of trucks that can be expected to travel between 34 and 50 thousand miles 0.4452 Proportion of trucks that can be expected to travel either less than 30 or more than 60 thousand miles P(x < 30)P(x > 60) Z-score for (x = 30): Z= 30-50 Z= 10 z=-2,00 Z-score for (x = 60): Z = 60-50 10 z= 1.00 Hence, P(x< 30) P(x > 60) P(z <-2.00) P(z > 1.00) From normal distribution table; P(z < -2.00) P(z > 1.00) 0.0228 0.1587 = 0.1814 Therefore, proportion of trucks that can be expected to travel either less than 30 or more than 60 thousand miles 0.1814

Value of z-score, such that area under the normal curve to the right of it equal to 80%, is given by: z =-0.842 1- x x -50 0.842 = 10 x 41.58 Hence, number of miles travelled by at least 80 percent of the trucks 41.58 thousand