Answer:
The confidence interval at at 99% level of confidence is <em>93.7 ≤ μ ≤ 96.9</em>.<em> </em>
Step-by-step explanation:
Step 1:
We must first determine the z-value at a confidence level of 99%.
Therefore,
99% = 100%(1 - 0.01)
Thus,
α = 0.01
Therefore, the z-value will be
z_(α/2) = z_(0.01/2) = z_0.005 = 2.58
(The z-value is read-off from the z table from the standard normal probabilities.)
Step 2:
We can now write the confidence interval:
X - z_(α/2) [s/√(n)] ≤ μ ≤ X + z_(α/2) [s/√(n)]
95.3 - 2.58(6.5/√(104)) ≤ μ ≤ 95.3 + 2.58(6.5/√(104))
<em>93.7 ≤ μ ≤ 96.9</em>
Therefore, confidence interval is <em>93.7 ≤ μ ≤ 96.9 </em>which means that we are 99% confident that the true mean population lies is at least 93.7 and at most 96.9.
Answer:
1.3
Step-by-step explanation:
26/3 is 1.3times larger than 20/3
Answer:
He wont foul out with probability 0.9093
Step-by-step explanation:
The total number of fools he picked is a Binomial ditribution noted by X with parameters p = 0.05 and N = 48. The mean of this random variable is μ = np = 48*0.05 = 2.4 and the variance is σ² = np(1-p) = 2.4*0.95 = 2.28, hence its standard deviation is σ = √2.28 = 1.51.
Note that, if approximate probability is asked, we could just approximate X with a Normal random variable with mean 2.4 and standard deviation 1.51 (this can be done because of the central limit theorem). We will calculate the probability manually. He wont foul out if he picks 0,1,2,3 or 4 fouls, thus
As a consecuence, he wont foul out with probability 0.9093.
Take 2,720/34,000=.0806. This means that 2,720 is 8.06% of the original number. In order to find the percentage of decrease, you take 100%-8.06%, which equals 91.94%. Your answer is 91.94%.
