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.
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.
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