Question

A particular bulb has lifetimes of 1500 hours with a standard deviation of 6 hours. Find a range in which it can be guaranteed that 93.75% of the lifetimes of this brand of car lie.

Answer 1

We are given,

$\mu=1500,\space \sigma=6$

To find the range, we are required to find the probability

$p(x_0\lt X\lt x_1)=0.9375$, where $x_0$ is the lower bound value and $x_1$ is the upper bound value.

We have to standardize since we are given the population mean $(\mu)$ and variance $(\sigma)$ as below

$p((x_0-\mu)/\sigma\lt (X-\mu)/\sigma \lt (x_1-\mu)/\sigma)=0.9375$

This can be written as,

$p((x_0-\mu)/\sigma\lt Z\lt (x_1-\mu)/\sigma)=0.9375$

Define,

$Z_0=(x_0-\mu)/\sigma$ and $Z_1=(x_1-\mu)/\sigma$

Now we have,

$p(Z_0\lt Z\lt Z_1)=0.9375$

The point $Z_0$ leaves an area of $(1-0.9375)/2=0.03125$ to the left and the point $Z_1$ leaves an area of $(1-0.9375)/2=0.03125$ to the right because of symmetry.

So,

$p(Z\lt Z_0)=0.03125$ and we can obtain the value of $Z_0$ from standard normal tables using the command $qnorm(0.03125)=-1.862732$ in $R$. Thus $Z_0=-1.862732$.

Since $Z_0=(x_0-\mu)/\sigma=-1.862732$, we can obtain the value of $x_0$ by substituting for $\mu$ and $\sigma$ as below,

$(x_0-1500)/6=-1.862732$

$x_0=(6*-1.862732)+1500=1488.82361$

The area to the left of $Z_1$ is $(1-0.03125)=0.96875$. We can express this as,

$p(Z\lt Z_1)=0.96875$. $Z_1$ can be obtained from the standard normal tables using the command $qnorm(0.96875)=1.862732$ in $R$ . Hence, $Z_1=1.862732$.

Since $Z_1=(x_1-\mu)/\sigma=1.862732$, the value of $x_1$ can be found by substituting for $\mu$ and $\sigma$ as,

$(x_1-1500)/6=1.862732$

$x_1=(6*1.862732)+1500=1511.17639$.

Therefore, $(1488.82, 1511.18)$ is a range in which it can be guaranteed that 93.75% of the lifetimes of this brand of car lie.