*X* = life of motor

*x* = guarantee period

We need to find the value (in years) that will give us the bottom 3% of the distribution. These are the motors that we are willing to replace under the guarantee.

`P(X < x) = 0.03`

The area that we can find from the *z*-table is

`0.5 - 0.03 = 0.47`

The corresponding *z*-score is `z = -1.88`.

Since `Z=(x-mu)/sigma`, we can write:

`(x-10)/2=-1.88`

Solving this gives `x = 6.24.`

So the guarantee period should be `6.24` years.

Here's a graph of our situation. Our normal curve has *μ* = 10, *σ* = 2.

The yellow portion represents the 47% of all motors that we found in the *z*-table (that is, between 0 and −1.88 standard deviations).

The light green portion on the far left is the 3% of motors that we expect to fail within the first 6.24 years.

The left-most portion represents the 3% of motors that we are willing to replace.

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