1. ## normally distributed lifespan

I have the general idea to approach this problem ; however, I don't understand it completely , so I would like to get some help.

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 10.1 years, and standard deviation of 1.8 years. The lowest 7% of items will last less than how many years?

2. Originally Posted by skorpiox
I have the general idea to approach this problem ; however, I don't understand it completely , so I would like to get some help.

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 10.1 years, and standard deviation of 1.8 years. The lowest 7% of items will last less than how many years?
You require the value of x* such that Pr(X < x*) = 0.07.

Get the value of z* such that Pr(Z < z*) = 0.07 (from tables or technology).

The it follows from $\displaystyle Z = \frac{X - \mu}{\sigma}$ that $\displaystyle z* = \frac{x* - 10.1}{1.8}$.

Substitute the value of z* and solve for x*.

3. Originally Posted by skorpiox
I have the general idea to approach this problem ; however, I don't understand it completely , so I would like to get some help.

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 10.1 years, and standard deviation of 1.8 years. The lowest 7% of items will last less than how many years?
Compute the Z-score for 0.07 using a table or the InvNormal function on the TI graphing calculator.

Translate this Z-score into X using:

$\displaystyle z = \frac{X - \mu}{\sigma}$

$\displaystyle \mu = 10.1$
$\displaystyle \sigma = 1.8$
X = unknown
z = known

My calculation indicates X = 7.4435. Can you confirm?

4. Thank you guys .By the way , Apcalculus your answer is correct