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Answer:
A. Initially, there were 12 deer.
B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.
C. After 15 years, there will be 410 deer.
D. The deer population incresed by 30 specimens.
Step-by-step explanation:
The amount of deer that were initally in the reserve corresponds to the value of N when t=0
A. Initially, there were 12 deer.
B.
B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.
C.
C. After 15 years, there will be 410 deer.
D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:
ΔN=N(15)-N(10)
ΔN=410 deer - 380 deer
ΔN= 30 deer.
D. The deer population incresed by 30 specimens.
Answer:
The probability is 1/2
Step-by-step explanation:
The time a person is given corresponds to a uniform distribution with values between 0 and 100. The mean of this distribution is 0+100/2 = 50 and the variance is (100-0)²/12 = 833.3.
When we take 100 players we are taking 100 independent samples from this same random variable. The mean sample, lets call it X, has equal mean but the variance is equal to the variance divided by the length of the sample, hence it is 833.3/100 = 8.333.
As a consecuence of the Central Limit Theorem, the mean sample (taken from independant identically distributed random variables) has distribution Normal with parameters μ = 50, σ= 8.333. We take the standarization of X, calling it W, whose distribution is Normal Standard, in other words
The values of the cummulative distribution of the Standard Normal distribution, lets denote it , are tabulated and they can be found in the attached file, We want to know when X is above 50, we can solve that by using the standarization