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 in a normal distribution, about 68% of the data lies within 1 standard deviation of the mean, 95% within 2 standard deviations, and 99.7% within 3 standard deviations. The area between $z=-3$ and $z = - 2$ can be found by considering the cumulative percentages. The area between $z=-3$ and $z = 3$ is 99.7%, the area between $z=-2$ and $z = 2$ is 95%. The area between $z=-3$ and $z=-2$ is $\frac{99.7 - 95}{2}=2.35%$.

Answer:

2.35%