Is 94.598 greater than 94.643?
1 answer:
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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
Answer:
Step-by-step explanation:
So we have:
Multiply both sides by 11:
The right side cancels:
Multiply the left:
Thus, the value of q is 33.
Answer:
No, there is not enough evidence at the 0.02 level to support the strategist's claim.
Step-by-step explanation:
We are given that a political study took a sample of 1300 voters in the town and found that 57% of the residents favored construction.
And, a political strategist wants to test the claim that the percentage of residents who favor construction is more than 54%, i.e;
Null Hypothesis, : p = 0.54 {means that the percentage of residents who favor construction is 54%}
Alternate Hypothesis, : p > 0.54 {means that the percentage of residents who favor construction is more than 54%}
The test statistics we will use here is;
T.S. = ~ N(0,1)
where, p = actual percentage of residents who favor construction = 0.54
= percentage of residents who favor construction in a sample of
1300 voters = 0.57
n = sample of voters = 1300
So, Test statistics =
= 2.185
Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have insufficient evidence to reject null hypothesis.
Therefore, we conclude that the percentage of residents who favor construction is 54% and the strategist's claim is not supported.