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#1 2009-07-30 13:01:59

finitehelp
Member
Registered: 2009-06-21
Posts: 80

Problem help!

pLEASE HELP IF YOU CAN...


1. Solve the problem.

The life span of a certain type of car timing belt, calculated in miles, is normally distributed, with a mean of 60,000 miles and a standard deviation of 7500 miles. If the maker of the timing belt wants less than 4% of the belts to fail while under warranty, for how many miles should the timing belts be guaranteed?


2.At one high school, students can run the 100-yard dash in an average of 15.2 seconds with a standard deviation of .9 seconds. The times are very closely approximated by a normal curve. Find the percent of times that are:

Between 17 and 17.9 seconds

3. Assume the distribution is normal. Use the area of the normal curve to answer the question. Round to the nearest whole percent.

A machine produces pencils with an average diameter of .30 inches and a standard deviation of .01 inches. What is the probability of a pencil with a diameter less than .285 inches?

Last edited by finitehelp (2009-07-30 13:52:58)

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#2 2009-07-30 19:28:21

riad zaidan
Guest

Re: Problem help!

#1: you find the value of z corresponding to .04 from the tables wcich is -1.75
therefore (x-60,000)/7500 =-1.75     x=46875 miles     wanted


#2           z1= (17-15.2)/.9=2
               z2=(17.9-15.2)/.9=3
the area between these values is .9987-.9772=.0215
the percent is =2.15%


#3 p(x<.285)=p(z<((0.285-0.3)/0.01)=p(z<-1.5)=.0559~.056

#3 2009-07-30 19:38:09

riad zaidan
Guest

Re: Problem help!

riad zaidan wrote:

#1: you find the value of z corresponding to .04 from the tables wcich is -1.75
therefore (x-60,000)/7500 =-1.75     x=46875 miles     wanted


#2           z1= (17-15.2)/.9=2
               z2=(17.9-15.2)/.9=3
the area between these values is .9987-.9772=.0215
the percent is =2.15%


#3 p(x<.285)=p(z<((0.285-0.3)/0.01)=p(z<-1.5)=.0559~.056

#4 2009-07-30 22:52:43

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Problem help!

#3 p(x<.285)=p(z<((0.285-0.3)/0.01)=p(z<-1.5)=.0559~.056

I am getting for P(z<-1.5) =.0668 ≈.067


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#5 2009-08-01 04:02:39

riad zaidan
Guest

Re: Problem help!

Hi
thats right sir
thank you
Riad Zaidan

#6 2009-08-01 07:21:21

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Problem help!

It's OK, that happens to me all the time.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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