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.
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
The new mean lifetime is
Explanation:
We are told
#"P"(X > 1480) = 0.98#
The way to solve this problem is to "translate"
#z = (x-mu)/sigma#
The rule is, if
For this question, we get
#"P"(X > 1480) = 0.98#
#=>"P"(Z > z)=0.98#
where
#"P"(Z < z)=0.02" "=> " "z ~~ –2.055#
That is, about 2% of the area under the
#" "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
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#