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il63 [147K]
4 years ago
6

What are the conditions a sample needs to meet before you can assume it's binomial and that it approximates a normal distributio

n?
Mathematics
1 answer:
Gemiola [76]4 years ago
8 0

Answer:

1) np\geq 5

2) nq = n(1-p)\geq 5

Other conditions that are important are:

3) n is large

4) p is close to 1/2 or 0.5

Step-by-step explanation:

1) Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

2) Solution to the problem

Let X the random variable of interest, on this case we now that:  

X \sim Binom(n, p)  

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

In order to apply the normal apprximation we need to satisfy these two conditions:

1) np\geq 5

2) nq = n(1-p)\geq 5

Other conditions that are important are:

3) n is large

4) p is close to 1/2 or 0.5

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Answer:

The width of the football field is 160 feet.

The length of the football field is 360 feet.

Step-by-step explanation:

Let w represent width of the football field.

We have been given that the length is 200 ft more than the width, so the length of the field would be w+200.

We are also told that the perimeter is 1,040 ft. We know that football field is in form of rectangle, so perimeter of field would be 1 times the sum of length and width. We can represent this information in an equation as:

2(w+w+200)=1040

Let us solve for w.

2(2w+200)=1040

4w+400=1040

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Upon substituting w=160 in expression w+200, we will get length of field as:

w+200\Rightarrow 160+200=360

Therefore, the length of the football field is 360 feet.

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