SOLUTION: Suppose that a tire manufacturer believes that the lifetimes of its tires follow a normal distribution with mean 48,000 miles and standard deviation 5,000 miles. 1. Produce a we

Question 967133: Suppose that a tire manufacturer believes that the lifetimes of its tires follow a normal distribution with mean 48,000 miles and standard deviation 5,000 miles.
1. Produce a well-labeled sketch of this normal distribution.
2. Determine the z-score corresponding to 55,000 miles.
3. Determine the probability that a randomly selected tire lasts for more than 55,000 miles.
4. If the manufacturer wants to issue a guarantee so that 99% of its tires last for longer than the guaranteed lifetime, what z-score should it use to determine that guaranteed lifetime?
5. If the manufacturer wants to issue a guarantee so that 99% of its tires last for longer than the guaranteed lifetime, how many miles should it advertise as its guaranteed lifetime?

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
# 1
Sketch

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# 2

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# 3

Mark 55,000 on the normal distribution sketch done in problem #1

Then shade to the right to represent the proportion, probability or percentage of getting a tire lasting more than 55,000 miles.

Using a calculator like a TI84 or something like this calculator, the probability is roughly 0.0807566592 (notice how it's the orange area under the curve to the right of 55,000)

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# 4

Use a calculator to compute the inverse cumulative distribution. Basically the reverse of what happened in #3

You can use the same calculator I provided to you, just click the "Value from an area (Use to compute Z for confidence intervals)" radio button to change the mode.

That calculator spits out z = -2.327, which is the answer to this problem.

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# 5

Use the z-score found in #4 to find the raw score

So they should advertise "99% of our tires last longer than 36,365 miles" (the other 1% last less than 36,365 miles)
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