Question

Let `X` be `N(mu, sigma^2)` so that `P(X<89) = 0.90` and `P(X<94) = 0.95`. How do you find `mu` and `sigma^2`?

Answer 1

Relate the given normal probabilities to their corresponding standard normal probabilities. Form a system of equations. Solve for the two unknowns.

`mu = 71.70," "sigma^2 = 182.52`.

Explanation:

We find the associated value of `z` from the standard normal distribution `Z" ~ "N(0, 1)`, such that `"P"(X < x) = "P"(Z < z)`, using the relation `z = (x - mu)/sigma`, as follows:

`"P"(X < x) = "P"(Z < (x - mu)/sigma)`

so

`"P"(X<89) = "P"(Z < (89 - mu)/sigma)`
`"            "0.90 = "P"(Z < (89 - mu)/sigma)`

And from `z`-table lookup, the `z` value that gives `"P"(Z < z)= 0.90` is `z = 1.28`. Thus, we know

`z = 1.28 = (89 - mu) / sigma`

Similarly, we also use `"P"(X< 94) = 0.95` to get

`z = 1.65=(94 - mu)/sigma`

We now have a system of two equations in two unknowns: `mu` and `sigma`. We can solve this system as follows:

`{(1.28 sigma = 89 - mu),(1.65 sigma = 94 - mu):}`

`=> {(mu = 89 - 1.28 sigma),(mu = 94 - 1.65 sigma):}`

`=>89 - 1.28 sigma = 94 - 1.65 sigma`

`=>"         "0.37 sigma = 5`

`=>"                "sigma = 13.51`

so

`mu = 89 - 1.28 sigma`

`=>mu = 89 - 1.28(13.51)`
`color(white)(=> mu) = 89 - 17.30`
`color(white)(=> mu) = 71.70`

Finally, `sigma = 13.51` means `sigma ^2 = 13.51^2 = 182.52`.

Thus, we have `mu = 71.70` and `sigma = 182.52`.