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
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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?
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)
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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