What proportion of tires will fail at 74000 mi or less? etc.

[lolohelp]

Hi everyone,
My name is Laurie, and I've been trying to solve a statistic question but without any success, so I was wondering if one of you is able to help me on 2 questions:

1- A certain brand of tires has a mean life of 80 and standard deviation of 6 (in thousands of miles).
a)-What proportion of tires will fail at 74 thousand miles or less?
b)-What should the warranty be if at most 5 percent of the tires would fail before or at the warrenty?

for 1a) I found P(x<74)= .1587, but it doesn't look right to me, and for 1b) I don't understand the question.
So, if you have any ideas, please let me know

and my second question is:

2- Two dogs out of 5 are chosen so that the first gets a treat and the second gets a root beer.
a)-How many outcomes are possible?

For this question, I still have no idea on how to do it.

Please let me know as soon as possible

thank you so much

Best Regards

Laurie

 

[tkhunny]

Re: statistic question

1- A certain brand of tires has a mean life of 80 and standard deviation of 6 (in thousands of miles).
a)-What proportion of tires will fail at 74 thousand miles or less?
for 1a) I found P(x<74)= .1587, but it doesn't look right to me, and for

What does "doesn't look right" mean? Is it or isn't it? Unique answers don't care if you don't like them.

74 is below 80, so the probability must be less than 0.50.
74 is exactly one standard deviation below the mean. What is the empirical rule for that? 0.50 - 0.34 = 0.17.
Is it looking any better, yet?

1b) I don't understand the question.
b)-What should the warranty be if at most 5 percent of the tires would fail before or at the warrenty?

This is the same question, but asked in reverse. Rather than give the value and ask the percentage, you have the percentage and are asked the value. 'x' is the value you seek.

Pr(z < 0.05) = -1.644854

(x-80)/6 = -1.644854

Find 'x'.

Please look at how these two problems relate to each other.

 

[lolohelp]

Re: statistic question

Dear Tkhunny,
So what you are saying is for question 1a) is:
I need to find the z-score so: z=(74000 - 80000)/6000 = -1
p(x < 74) = P(z<-1)= .3413 .5 - .3413=.1587
a).1587

and for 1b), the answer will be:
z<0.05 = -1.64
(x-80)/6 = -1.64

x-80=-1.64 X 6
x-80= -9.84
x= 70.16
Is that Right?
b) 70.16

 

[tkhunny]

Re: statistic question

I get 70.13088, but I probably used less rounding. Good work.