How do you find the new mean for a normal distribution, given standard deviation and probability?

A company manufactures batteries with a lifetime which is normally distributed with a mean of 1500 and a standard deviation of 70 hours.
The company decides to improve the mean lifetime of the batteries, but retain the same standard deviation. If 98% of all batteries last more than 1480 hours, determine the new mean lifetime.

1 Answer
Jul 24, 2018

The new mean lifetime is #mu=1623.85# hours.

Explanation:

We are told #X# is normally distributed with #sigma=70#, and we are given

#"P"(X > 1480) = 0.98#

The way to solve this problem is to "translate" #X# to the standard normal distribution #Z# by using

#z = (x-mu)/sigma#

The rule is, if #X# is normally distributed with mean #mu# and standard deviation #sigma#, then for any point #x# along the distribution of #X#, there is a "partner" point #z# on the standard normal curve #Z# such that #"P"(X > x) = "P"(Z > z).# That partner point #z# is equal to #(x-mu)/sigma.#

For this question, we get

#"P"(X > 1480) = 0.98#

#=>"P"(Z > z)=0.98#

where #z = (x-mu)/sigma = (1480-mu)/70#. Now, we use the #z#-table to reverse-lookup the #z#-coordinate of #Z# that has an area to its right of 0.98 (in other words, an area to its left of 0.02). By table lookup, we find

#"P"(Z < z)=0.02" "=> " "z ~~ –2.055#

That is, about 2% of the area under the #Z# curve is to the left of the point #z=–2.055#. And since #z = (x - mu)/sigma,# we can now solve for #mu:#

#"            "z = (x - mu)/sigma#

#"  "–2.055 = (1480 - mu)/70#

#–143.85 = 1480 - mu#

#"            "mu = 1480 + 143.85#

#"            "mu = 1623.85#

So, the new mean is #mu=1623.85#.
#" "#
#" "#

Summary:

#"P"(X > 1480) = 0.98#

#=>"P"(Z > (1480 - mu)/70) = 0.98#

#=>(1480-mu)/70 = –2.055#

#=>1480-mu = –143.85#

#=>"             "mu = 1623.85#