A survey of 120 college junior business students produced these results: 40 read a national newspaper, 48 read the local paper,7
1 answer:
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Let denote the value on the
-th drawn ball. We want to find the expectation of
, which by linearity of expectation is
(which is true regardless of whether the are independent!)
At any point, the value on any drawn ball is uniformly distributed between the integers from 1 to 10, so that each value has a 1/10 probability of getting drawn, i.e.
and so
Then the expected value of the total is
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 ≥ )
⇒ 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