Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = 3 (*sqaure root sign*) t (20 − t), [0,
1 answer:
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Answer:
- maximum: 15∛5 ≈ 25.6496392002
- minimum: 0
Step-by-step explanation:
The minimum will be found at the ends of the interval, where f(t) = 0.
The maximum is found in the middle of the interval, where f'(t) = 0.
This derivative is zero when the numerator is zero, at t=5. The function is a maximum at that point. The value there is ...
f(5) = (∛5)(20-5) = 15∛5
The absolute maximum on the interval is 15∛5 at t=5.
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Answer:
(1) A Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.
(2) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.
(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.
Step-by-step explanation:
Consider a random variable <em>X</em> following a Binomial distribution with parameters <em>n </em>and <em>p</em>.
If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:
- np ≥ 10
- n(1 - p) ≥ 10
The three populations has the following proportions:
p₁ = 0.10
p₂ = 0.30
p₃ = 0.50
(1)
Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:
Thus, a Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.
(2)
Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:
Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.
(3)
Check the Normal approximation conditions for population 3, for all the provided <em>n</em> as follows:
Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.