Question

Joseph Mart recently purchased new trucks for its delivery fleet. The truck manufacturer confirmed

that the engines in the trucks have a mean failure time of 8 years, with a standard deviation of

1.5 years. Within what amount of time should Walmart expect for 93.32% of the trucks to fail?

Please show all work, with a normal distribution curve if possible, thanks in advance!

Answer 1

Solution:

Given: the trucks have a mean failure time of 8 years, with a standard deviation of 1.5 years.

That is : Mean =
u= 8
and standard deviation =
0 = 1.5

We have to find the values of x such that:

P( x₁ < X < x₂ ) = 93.32%

P( x₁ < X < x₂ ) = 0.9332

Thus find z values such that:

P( -z  < Z < z  ) = 0.9332

If area between -z to +z is 0.9332 , then area outside this interval is = 1 - 0.9332 = 0.0668

That is area in tails is 0.0668

We divide this area in two tails equally

that is: 0.0668 / 2 = 0.0334 in left and right tail.

That is:

P( Z< -z) = 0.0334  

and

P( Z> z )= 0.0334

Thus we get:

Pl-z<Z<z )= 0.9332 P(Z > z) = 0.0334 P(Z <-z) = 0.0334

Thus from above graph we can see total area below z is:

P( Z< z ) = P( Z< -z ) + P( -z < Z < z )

P( Z< z ) = 0.0334 + 0.9332

P( Z< z ) = 0.9666

Thus look in z table for area = 0.9666 or its closest area and find z value.

10.6 1.00 .01 .02 .4 .05 .06 .07 .08 .09] | 0.0 - 1,5000 5040 5080 .50 5160 .5199 .5239 5279 5319 .5359 | 10.1 ,5398 ,5438 .5478 . 57.5555596 .5636 .5675 15714 753 | 0.2 57935832 5871 .5910 5948 5987 .6026 .6064 .6103 .6141 | 0.3 6179 ,6217 1.6255 .623 .63316368 6406 6443 640 ,6517 .6554 6591 .6628664 6700 .5736 6772 .6808 ,684 .6879 | 0.5 6915 16950 054 70887123 71771907224 72577291 7324 37 389 7422.7454748675177549 | 0.7 7580 7611 ,7642 .768 04 74 7764 7794 78237852 | 10.8 788179107939797 1995 .8023 .8051 ,80788106 .8133 10.9 1.8159 ,8186 .8212 828 18264 .8289 8315 18340 ,8365 839| | 1.0 8413 .8438 ,8461 .84年 1.8508 .8531 1.8554 .8577 .8599 | 1.1 8643 ,8665 1.8729 8749 870 .8790 8810 | 12 .889 ,8869 ,888 .89T 1.8925894 1.8952 ,8980 ,8997 ,9015 | | 139032909906690329099 9115 913191479162 917 149192 9207 192229286 9251 9265 1979.9292 9306 9319] | 1.5 9332 9345 937 9 0 9382 9394 9406 9418 194299441| | 1. 6945294639474949495 9505 9515952595359545 | 1.7 9554 9564 9573 9962 19591 .9599 9608 19616 -9625 19633 1.8 -419640 0 .966496719678 196869693 9699 .9706 | 18621 18830

Area 0.9664 is closest to 0.9666 and it corresponds to 1.8 and 0.03

thus z = 1.83

thus -z = -1.83

Now for finding x value we use formula:

= μ +Χσ

T1 = 1+(-x) X Ở

I1 = 8+(-1.83) x 1.5

-21 = 8 -2.745

21 = 5.255

-21 = 5.3

and

T2 = μ+ (+2) xσ

22 = 8+1.83 x 1.5

C2 = 10.745

12 = 10.7

Thus within 5.3 years to 10.7 years   should Walmart expect for 93.32% of the trucks to fail.

P(5.3 < X < 10.7) = 0.9332 -2 -1.83 x1 = 5.3 years mean=8 years z = 1.83 x2 = 10.7 years