Probability: tire-lifetime has mean of 42100 mi, st. dev. of

[fyr]

A maker of tires for cars has a lot of evidence to support the fact that the lifetime of their tires follows a normal distribution with a mean of 42,100 miles and a standard deviation of 2,510 miles. Find the probability that a randomly selected tire will have a lifetime of between 44,500 miles and 48,000 miles. Be certain that you round your z-values to two decimal places. Round your answer to 4 decimal places.
Choose one answer.

a. 0.1685

b. 0.8315

c. 0.1591

d. 0.3315

e. None of these are correct.

z=44500-42100/2510=0.9562 probability using the z-table= .8292
z=48000-42100/2510=2.3506 prob=.9893

.8292-.9893=.1601

But this answer is not one of the choices. The professor told us that some of our numbers may be a little off due to rounding. So, before I choose E none of these are correct...is it possible the answer could be 0.1685 or 0.1591?

Or maybe I did the calculations wrong? Please help me. I will be so grateful!

 

[tkhunny]

Re: Probability

1) Notation needs work. "44500-42100/2510" doesn't mean what you want. Use "(44500-42100)/2510".

2) You didn't follow the instructions. You should have Z1 = 0.96 and Z2 = 1.35 Giving 0.9115-0.8315

3) Your subtraction is backwards. 0.8292 - 0.9893 is positive?

4) .9893 relates to Z = 2.30. How did you get that?

Just give it another try. Be more careful.