a company claims the life of its car batteries is normally distributed with a mean of 3.1 years and a…

a company claims the life of its car batteries is normally distributed with a mean of 3.1 years and a standard deviation of 0.45 years. using the 68 - 95 - 99.7% rule, what percentage of the companys car batteries will last between 1.75 and 2.2 years? enter the answer in the box.

a company claims the life of its car batteries is normally distributed with a mean of 3.1 years and a standard deviation of 0.45 years. using the 68 - 95 - 99.7% rule, what percentage of the companys car batteries will last between 1.75 and 2.2 years? enter the answer in the box.

Answer

Explanation:

Step1: Calculate z - scores

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu = 3.1$ (mean), $\sigma=0.45$ (standard deviation). For $x = 1.75$, $z_1=\frac{1.75 - 3.1}{0.45}=\frac{- 1.35}{0.45}=-3$. For $x = 2.2$, $z_2=\frac{2.2 - 3.1}{0.45}=\frac{-0.9}{0.45}=-2$.

Step2: Use the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that about 99.7% of the data lies within 3 standard deviations of the mean ($z=-3$ to $z = 3$) and about 95% of the data lies within 2 standard deviations of the mean ($z=-2$ to $z = 2$). The percentage of data between $z=-3$ and $z=-2$ is $\frac{99.7 - 95}{2}=2.35%$.

Answer:

$2.35$