How Do You Solve for a in a Normal Distribution Given Probability Ratios?

[Of Mike and Men]

Homework Statement

Suppose that X ~ N(μ,σ). Find a in terms of μ and σ if P(X>a) = 1/3 * P(X ≤a)

Homework Equations

The Attempt at a Solution

1 - P(X ≤a) = 1/3 * P(X ≤a)
3 = 4P(X ≤a)
P(X ≤a) = 3/4

Since x₀ = μ + σz₀ where x₀ and z₀ are the same percentile for N(μ,σ) and N(0,1) (respectively), then z₀ = 0.67449 (by invNorm(0.75, 0, 1)). It follows that

a = μ + 0.67449σ

I'm not sure if this is the correct method, someone in my class solved it another way and got a different answer, but this seems to make sense to me.

Regards,

Michael

 

[Charles Link]

I agree with your solution. Also please read my last comments on your Poisson distribution problem of last week. I think you'll find them of interest. That was actually the first time I had looked at the Poisson distribution in this much detail, and the two methods of solving the same problem are both quite interesting.

 

[Ray Vickson]

Homework Statement

Suppose that X ~ N(μ,σ). Find a in terms of μ and σ if P(X>a) = 1/3 * P(X ≤a)

Homework Equations

The Attempt at a Solution

1 - P(X ≤a) = 1/3 * P(X ≤a)
3 = 4P(X ≤a)
P(X ≤a) = 3/4

Since x₀ = μ + σz₀ where x₀ and z₀ are the same percentile for N(μ,σ) and N(0,1) (respectively), then z₀ = 0.67449 (by invNorm(0.75, 0, 1)). It follows that

a = μ + 0.67449σ

I'm not sure if this is the correct method, someone in my class solved it another way and got a different answer, but this seems to make sense to me.

Regards,

Michael

It is the correct method, and the answer is correct.