Question

A trucking company determined that the distance traveled per truck per year is normally? distributed, with a mean of 50 thousand miles and a standard deviation of 10 thousand miles. Complete parts? (a) through? (c) below. A. What proportion of trucks can be expected to travel between 37 and 50 thousand miles in a? year? The proportion of trucks that can be expected to travel between 37 and 50 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 40 or more than 70 thousand miles in a? year? The percentage of trucks that can be expected to travel either less than 40 or more than 70 thousand miles in a year is _ %. ?(Round to two decimal places as? needed.) c. How many miles will be traveled by at least 65?% of the? trucks? The amount of miles that will be traveled by at least 65?% of the trucks is _ miles. ?(Round to the nearest mile as? needed.)

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	50
std deviation   =σ=	10.0000

probability =

P(37<X<50)

=

P(-1.3<Z<0)=

0.5-0.0968=

0.4032

b)

probability =

1-P(40<X<70)

=

1-P(-1<Z<2)=

1-(0.9772-0.1587)=

1-0.8185=0.1815~ 18.15%

c)

for 35th percentile critical value of z=		-0.39
therefore corresponding value=mean+z*std deviation=			46.10~ 46

hence 46 miles will be traveled by at least 65 % of the truck