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Alex17521 [72]
3 years ago
9

"is it appropriate to use the normal approximation for the sampling distribution of"

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
5 0

Answer: Normal approximation can be used for discrete sampling distributions, such as Binomial distribution and Poisson distribution if certain conditions are met.

Step-by-step explanation: We will give conditions under which the Binomial and Poisson distribitions, which are discrete, can be approximated by the Normal distribution. This procedure is called normal approximation.

1. Binomial distribution: Let the sampling distribution be the binomial distribution B(n,p), where n is the number of trials and p is the probability of success. It can be approximated by the Normal distribution with the mean of np and the variance of np(1-p), denoted by N(np,np(1-p)) if the following condition is met:

n>9\left(\frac{1-p}{p}\right)\text{ and } n>9\left(\frac{p}{1-p}\right)

2. Poisson distribution: Let the sampling distribution be the Poisson distribution P(\lambda) where \lambda is its mean. It can be approximated by the Normal distribution with the mean \lambda and the variance \lambda, denoted by N(\lambda,\lambda) when \lambda is large enough, say \lambda>1000 (however, different sources may give different lower value for \lambda but the greater it is, the better the approximation).

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Irina-Kira [14]

Answer:

for a  a) {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}

\left[\begin{array}{cccc}0&1&1&1\\0&0&1&1\\0&0&0&1\\0&0&0&0\end{array}\right]

for b

b) {(1, 1), (1, 4), (2, 2), (3, 3), (4, 1)}

\left[\begin{array}{cccc}1&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{array}\right]

for c) {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

\left[\begin{array}{cccc}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{array}\right]

for d d) {(2, 4), (3, 1), (3, 2), (3, 4)}

\left[\begin{array}{cccc}0&0&0&0\\0&0&0&1\\1&1&0&1\\0&0&0&0\end{array}\right]

Step-by-step explanation:

in matrix, arrays are placed in rows , which represents the horizontal sides from left to right, while arrays in the column are placed vertically from top to bottom. Here, we placed the arrays in a 4x4 matrix

for a  a) {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}

\left[\begin{array}{cccc}0&1&1&1\\0&0&1&1\\0&0&0&1\\0&0&0&0\end{array}\right]

for b

b) {(1, 1), (1, 4), (2, 2), (3, 3), (4, 1)}

\left[\begin{array}{cccc}1&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{array}\right]

for c) {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

\left[\begin{array}{cccc}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{array}\right]

for d d) {(2, 4), (3, 1), (3, 2), (3, 4)}

\left[\begin{array}{cccc}0&0&0&0\\0&0&0&1\\1&1&0&1\\0&0&0&0\end{array}\right]

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3 years ago
Cuantas unidades mide al lado de pqrs
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to ketha ay ayu  uhh uh la toca bella

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3 years ago
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Answer with figure if needed.random answer will be reported.​
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10m/s

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C

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At any instant x2+y2=52

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8 0
3 years ago
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mariarad [96]

Answer:

\boxed{x = 4}

Step-by-step explanation:

=> 5x-4+2(x-4) = 16

Expanding the brackets

=> 5x-4+2x-8 = 16

Combining like terms

=> 5x+2x-4-8 = 16

=> 7x - 12 = 16

Adding 12 to both sides

=> 7x = 16+12

=> 7x = 28

Dividing both sides by 7

=> x = 4

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3 years ago
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Number of ice cream produced after = 54

<h2>Find:</h2>

Percent of speed the machine increase

<h2>Computation:</h2>

Percent of speed the machine increase = [Number of ice cream produced after - Number of ice cream produced before] / Number of ice cream produced before

Percent of speed the machine increase = [(54 - 45) / 45]100

Percent of speed the machine increase = [9 / 45]100

Percent of speed the machine increase = 20%

<h2>Learn more:</h2>

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2 years ago
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