Subtract the following. Write your answer in standard notation . 18-9.2×10^-4

Mathematics

1 answer:

puteri [66]2 years ago

8 0

Answer:1.799908×10^1

Step-by-step explanation:

18-0.00092

=17.99908

In standard form=1.799908×10^1

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

Katyanochek1 [597]

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).