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mote1985 [20]
3 years ago
15

The monthly salary of all adults in a certain US town is known to be right skewed with a mean of $3700. If a random sample of si

ze 40 adults were selected from that town, what would be the shape of the distribution of the sample mean?
Mathematics
1 answer:
Alexxandr [17]3 years ago
7 0

Answer:

The shape of the distribution of the sample mean would be bell-shaped(normally distributed).

Step-by-step explanation:

The Central Limit Theorem answers this question.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size, of at least 30, can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

In this problem, we have that

The population distribution is right-skewed.

Sampling distribution of the sample mean with size 40.

The Central Limit Theorem estabilishes that the sampling distribution of the sample mean will be normally distributed, even if the distribution of the population is not.

So the shape of the distribution of the sample mean would be bell-shaped(normally distributed).

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Simplify the expression and show work 4(3x-2)-7(x+5)
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Answer:

5x - 43

Explanation:

Given the below expression;

4\left(3x-2\right)-7\left(x+5\right)

To simplify the above, we'll go ahead and use the numbers outside the parentheses to multiply the ones inside to clear the parentheses as seen below;

\begin{gathered} 12x-8-7x-35 \\ =12x-7x-8-35 \\ =5x-43 \end{gathered}

So the answer is 5x - 43

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11 months ago
Rachel runs 3.2 miles each weekday and 1.5 miles each day of the weekend how many miles will she have run in six weeks
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3 years ago
0.5 grams of salt is equal to how many milligrams show you work
elena55 [62]

Answer:

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Step-by-step explanation:

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6 0
3 years ago
Read 2 more answers
A random sample of 5120 permanent dwellings on an entire reservation showed that 1627 were traditional hogans.
krek1111 [17]

Answer:

a) \hat p=\frac{1627}{5120}=0.3178

b)The 99% confidence interval would be given by (0.301;0.355)

Step-by-step explanation:

1) Notation and definitions

X=1627 number of permanent dwellings on the entire reservation that are traditional hogans.

n=5120 random sample taken

\hat p=\frac{1627}{5120}=0.318 estimated proportion of permanent dwellings on the entire reservation that are traditional hogans.

p true population proportion of permanent dwellings on the entire reservation that are traditional hogans.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

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

Part a

The point of estimate for p=the proportion of all permanent dwellings on the entire reservation that are traditional hogans is given by:

\hat p=\frac{1627}{5120}=0.3178

Part b

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.318 - 2.58\sqrt{\frac{0.318(1-0.318)}{5120}}=0.301

0.318 + 2.58\sqrt{\frac{0.318(1-0.318)}{5120}}=0.335

The 99% confidence interval would be given by (0.301;0.355)

8 0
3 years ago
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