Question

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.

Answer 1

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`