Answer to Question #264483 in Statistics and Probability for avi

Question #264483

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.


1
Expert's answer
2021-11-12T08:19:46-0500

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.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS