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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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PLEASE HELP ME The cost, C, to produce b baseball bats per day is modeled by the function C
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The minimum cost is the y-coordinate of the vertex of the parabola represented by the equation. The x-coordinate is the number of bats.

C(b)=0.06b^2-7.2b+390 \\ a=0.06 \\ b=-7.2 \\ \\
\hbox{the x-coordinate of the vertex: } \frac{-b}{2a}=\frac{-(-7.2)}{2 \times 0.06}=\frac{7.2}{0.12}=60


60 bats should be produced every day to keep costs at a minimum.
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Answer:

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Two functions, P and Q, are described as follows: Function P y = 5x + 3 Function Q The rate of change is 2 and the y-intercept i
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Consider a sample with data values of 27, 24, 21, 16, 30, 33, 28, and 24. Compute the 20th, 25th, 65th, and 75th percentiles. 20
densk [106]

Answer:

P_{20} = 20 --- 20th percentile

P_{25} = 21.75  --- 25th percentile

P_{65} = 27.85   --- 65th percentile

P_{75} = 29.5   --- 75th percentile

Step-by-step explanation:

Given

27, 24, 21, 16, 30, 33, 28, and 24.

N = 8

First, arrange the data in ascending order:

Arranged data: 16, 21, 24, 24, 27, 28, 30, 33

Solving (a): The 20th percentile

This is calculated as:

P_{20} = 20 * \frac{N +1}{100}

P_{20} = 20 * \frac{8 +1}{100}

P_{20} = 20 * \frac{9}{100}

P_{20} = \frac{20 * 9}{100}

P_{20} = \frac{180}{100}

P_{20} = 1.8th\ item

This is then calculated as:

P_{20} = 1st\ Item +0.8(2nd\ Item - 1st\ Item)

P_{20} = 16 + 0.8*(21 - 16)

P_{20} = 16 + 0.8*5

P_{20} = 16 + 4

P_{20} = 20

Solving (b): The 25th percentile

This is calculated as:

P_{25} = 25 * \frac{N +1}{100}

P_{25} = 25 * \frac{8 +1}{100}

P_{25} = 25 * \frac{9}{100}

P_{25} = \frac{25 * 9}{100}

P_{25} = \frac{225}{100}

P_{25} = 2.25\ th

This is then calculated as:

P_{25} = 2nd\ item + 0.25(3rd\ item-2nd\ item)

P_{25} = 21 + 0.25(24-21)

P_{25} = 21 + 0.25(3)

P_{25} = 21 + 0.75

P_{25} = 21.75

Solving (c): The 65th percentile

This is calculated as:

P_{65} = 65 * \frac{N +1}{100}

P_{65} = 65 * \frac{8 +1}{100}

P_{65} = 65 * \frac{9}{100}

P_{65} = \frac{65 * 9}{100}

P_{65} = \frac{585}{100}

P_{65} = 5.85\th

This is then calculated as:

P_{65} = 5th + 0.85(6th - 5th)

P_{65} = 27 + 0.85(28 - 27)

P_{65} = 27 + 0.85(1)

P_{65} = 27 + 0.85

P_{65} = 27.85

Solving (d): The 75th percentile

This is calculated as:

P_{75} = 75 * \frac{N +1}{100}

P_{75} = 75 * \frac{8 +1}{100}

P_{75} = 75 * \frac{9}{100}

P_{75} = \frac{75 * 9}{100}

P_{75} = \frac{675}{100}

P_{75} = 6.75th

This is then calculated as:

P_{75} = 6th + 0.75(7th - 6th)

P_{75} = 28 + 0.75(30- 28)

P_{75} = 28 + 0.75(2)

P_{75} = 28 + 1.5

P_{75} = 29.5

7 0
3 years ago
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