How Far from the Target Does a Machine Gun Hit 50% of Its Bullets?

[diracdelta]

Homework Statement

[/B]
While shooting at target, soldier shoots from machine gun fixed in a way so it can rotate only around z axis.
deviation from center of target due to twitching of machine gun is random variable with Gauss distribution.
Deviations are such that inside of 20m from center of target, it hits 80% bullets.
Find the distance, at how long away from center of target does 50% of the bullets end up.

The Attempt at a Solution

Since I'm dealing with gauss distribution, I must find σ and μ .
My problem is next. Idea is to integrate some parts and then i get numbers in table.
But what bothers me is next, what does 20 m and unknown distance represents?
What is σ and μ in this very problem?

 

[SteamKing]

Homework Statement

[/B]
While shooting at target, soldier shoots from machine gun fixed in a way so it can rotate only around z axis.
deviation from center of target due to twitching of machine gun is random variable with Gauss distribution.
Deviations are such that inside of 20m from center of target, it hits 80% bullets.
Find the distance, at how long away from center of target does 50% of the bullets end up.

The Attempt at a Solution

Since I'm dealing with gauss distribution, I must find σ and μ .
My problem is next. Idea is to integrate some parts and then i get numbers in table.
But what bothers me is next, what does 20 m and unknown distance represents?
What is σ and μ in this very problem?

When the gun is located 20 m from the center of the target, 80% of the bullets fired hit the target. Presumably, if the gun were located closer to the target than 20 m, the percentage of hits would be greater than 80%. If the gun were located say 10 cm away from the target, the percentage of hits would be 99.9999999...%.

What this problem is asking, if the gun were moved back from its original distance of 20 m from the target, how far from the target could the gun be and still have 50% of its shots hit the target.

 

[diracdelta]

Ok. I see.
Only idea i have is to find value in which integral of ((20-μ)/σ) gives 80%.
And if I set the same like ((20+x-μ)/σ) = 50%
I still cannot find σ or μ.

 

[diracdelta]

Ok, so i consulted with class asistent, he told me that we can freely assume μ=0. Knowing that, here's what I've come up with.

->Draw Gauss distribution.
Since inside 20 meteres probability is 80% i conclude Φ(20/σ) - Φ(-20/σ)=0.8
Due to simetric of problem, left and right flanks outside 80% area are both 0.1 or 10%, so Φ(-20/σ)=0.1 which yields Φ(20/σ)=0.9
(-20/σ)=1.28
σ=15.625

While knowing Φ(x/σ)=0.5
(x/σ)=0
So, x=0.

Do you think this is good?

 

[SteamKing]

Ok, so i consulted with class asistent, he told me that we can freely assume μ=0. Knowing that, here's what I've come up with.

->Draw Gauss distribution.
Since inside 20 meteres probability is 80% i conclude Φ(20/σ) - Φ(-20/σ)=0.8
Due to simetric of problem, left and right flanks outside 80% area are both 0.1 or 10%, so Φ(-20/σ)=0.1 which yields Φ(20/σ)=0.9
(-20/σ)=1.28
σ=15.625

While knowing Φ(x/σ)=0.5
(x/σ)=0
So, x=0.

Do you think this is good?

Obviously, if the gun stays where it already is, you will obtain 80% hits, and the last time I checked, 50% is still less than 80%, so that remains within the realm of possibility.

However, I think this problem wants you to find out how much farther away from the target than 20 meters can the gun be placed and still obtain 50% hits.