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iren2701 [21]
2 years ago
14

For a population with an unknown distribution, the form of the sampling distribution of the sample mean is _____. Group of answe

r choices approximately normal for small sample sizes exactly normal for large sample sizes exactly normal for small sample sizes approximately normal for large sample sizes
Mathematics
1 answer:
Nesterboy [21]2 years ago
6 0

For a population where the distribution is unknown, the sampling distribution of the sample mean will be b which is approximately normal for all sample sizes.

Given options regarding sample sizes.

We have to choose the correct option which shows the sample mean characteristics when the distribution is unknown.

Based on the central limit theorem the sampling distribution of the sample mean for either skewed variable ir normally distributed variable can be approximated given the mean and standard deviation.

The central limit theorem is said to be  true when n greater than or equal to 30.

When n is greater than 30 ,z test is used, when n is less than 30 ,t test is used.

Hence for a population where the distribution is unknown ,the sampling distribution of the sample mean will be approximately normal for all sample sizes.

Learn more about sample at brainly.com/question/24466382

#SPJ4

Question is incomplete as it includes options in a right way:

1)exactly normal for large sample,

2) approximately normal for all sample sizes,

3) exactly normal for all sample sizes.

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A new sample of 225 employed adults is chosen. Find the probability that less than 7.1% of the individuals in this sample hold m
arsen [322]

Answer:

The probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

Step-by-step explanation:

Let <em>X</em> = number of individuals in the United States who held multiple jobs.

The probability that an individual holds multiple jobs is, <em>p</em> = 0.13.

The sample of employed individuals selected is of size, <em>n</em> = 225.

An individual holding multiple jobs is independent of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 225 and <em>p</em> = 0.13.

But since the sample size is too large Normal approximation to Binomial can be used to define the distribution of proportion <em>p</em>.

Conditions of Normal approximation to Binomial are:

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Check the conditions as follows:

np=225\times 0.13=29.25>10\\n(1-p)=225\times (1-0.13)=195.75>10

The distribution of the proportion of individuals who hold multiple jobs is,

p\sim N(p, \frac{p(1-p)}{n})

Compute the probability that less than 7.1% of the individuals in this sample hold multiple jobs as follows:

P(p

*Use a <em>z</em>-table.

Thus, the probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

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Step-by-step explanation:

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Step-by-step explanation:

(2x(3+5)+6)(x4)

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