# Help with Statistics Homework? 10 points!!!?

Ok so here's the question:

A tire manufacturer designed a new tread pattern for its all-weather tires. Repeated tests were conducted on cars of approximately the same weight traveling at 60 miles an hour. The tests showed that the new tread pattern enables the cars to stop completely in an average distance of 125 feet with a standard deviation of 6.5 feet and that the stopping distances are approximately normally distributed.

a) What is the 70th percentile of the distribution of stopping distances? (I already got this one solved. I got 128.409 ft; feel free to check)

b) What is the probability that at least 2 cars out of 5 randomly selected cars in the study will stop in a distance that is greater than the distance calculated in part (a)? (this is the one I really need help with)

c) What is the probability that a randomly selected sample of 5 cars in the study will have a mean stopping distance of at least 130 feet?

Ive been trying at these problems for days and can't get b and c. Please help with as much as you can. I dont necessarily need the answers, just help with how to solve it. If you do give the answer, please let me know how you got it! Thanks sooo much for your time! No rude comments please! :) Thannkk youu <3

### 1 Answer

- BeeFreeLv 78 years agoFavorite Answer
Sungirl

Yes, answer A is correct :)

b) What is the probability that at least 2 cars out of 5 randomly selected cars in the study will stop in a distance that is greater than the distance calculated in part (a)? (this is the one I really need help with)

Well, the probability of a distance "greater" than 128.409 ft will be 1 - 0.7 = 0.3.

Now use Binomial ...

P(at least 2) = 1 - P(0) - P(1)

= 1 - 5C0 (0.3)^0 (0.7)^5 - 5C1 (0.3)^1 (0.7)^4 = 0.47178

c) What is the probability that a randomly selected sample of 5 cars in the study will have a mean stopping distance of at least 130 feet?

P(X > 130) = P[ z > (130-125) / (6.5 / sqrt 5)] = P(z > 1.72), now use Standard Normal table ...

P(z > 1.72) = 1 - P(z < 1.72) = 1 - 0.0427 = 0.9573

hope that helped