Word problem with mean and standard deviation

J

JellyFish

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Jan 12, 2009
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A distributor of auto tires claims that 90% of its best brand of tires last more than 140, 000 Km. It has been shown that the lifetime of the tires of that brand has a normal distribution with mean 100, 000 Km and a standard deviation of 20, 000 Km. Is the vendor's claim accurate?

We have done very basic problems like this in class but none this extensive.

From this I know that ? = 100, 000 and ? = 20,000.

I know I need to set up an inwquality with

a ? [(x-100,000)/20,000] ? b

but I don't know what a and b are to be and where the 140,000 and 90% fit in?

Thank you
 
JellyFish said:
A distributor of auto tires claims that 90% of its best brand of tires last more than 140, 000 Km. It has been shown that the lifetime of the tires of that brand has a normal distribution with mean 100, 000 Km and a standard deviation of 20, 000 Km. Is the vendor's claim accurate?

We have done very basic problems like this in class but none this extensive.

From this I know that ? = 100, 000 and ? = 20,000.

I know I need to set up an inwquality with

a ? [(x-100,000)/20,000] ? b

but I don't know what a and b are to be and where the 140,000 and 90% fit in?

Thank you
Click to expand...

Use z-distribution table.

But you should be able to tell this one without z-table.
 
That's what I want to do but I don't know what to do with the 140,000 and 90%?
 
Oh I figured it out using the Z score and solving for X, which I foun to be less than 140,000 so the vendor was false.


Thank you!
 
To have a success rate (minimum) greater than 50% - the claim must be below the "normal" mean.
 
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