Answer:
the probability that we hit the bullseye at least 100 times is 0.0113
Step-by-step explanation:
Given the data in the question;
Binomial distribution
We find the probability of hitting the dart on the disk
⇒ Area of small disk / Area of bigger disk
⇒ πR₁² / πR₂²
given that; disk-shaped board of radius R² = 5, disk-shaped bullseye with radius R₁ = 1
so we substitute
⇒ π(1)² / π(5)² = π/π25 = 1/25 = 0.04
Since we have to hit the disk 2000 times, we represent the number of times the smaller disk ( BULLSEYE ) will be hit by X.
so
X ~ Bin( 2000, 0.04 )
n = 2000
p = 0.04
np = 2000 × 0.04 = 80
Using central limit theorem;
X ~ N( np, np( 1 - p ) )
we substitute
X ~ N( 80, 80( 1 - 0.04 ) )
X ~ N( 80, 80( 0.96 ) )
X ~ N( 80, 76.8 )
So, the probability that we hit the bullseye at least 100 times, P( X ≥ 100 ) will be;
we covert to standard normal variable
⇒ P( X ≥ [tex]\frac{100-80}{\sqrt{76.8} }[/tex] )
⇒ P( X ≥ 2.28217 )
From standard normal distribution table
P( X ≥ 2.28217 ) = 0.0113
Therefore, the probability that we hit the bullseye at least 100 times is 0.0113
