Mathematics, 01.07.2020 15:01 lmiranda5034

The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return. The mean amount of deductions for this population of taxpayers was $16,642. Assume the standard deviation is σ = $2,400. a. What is the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for each of the following sample sizes: 20, 50, 100, and 500? (Round your answers to four decimal places.)
b. What is the advantage of a larger sample size when attempting to estimate the population mean?
a. A larger sample increases the probability that the sample mean will be a specified distance away from the population mean.
b. A larger sample increases the probability that the sample mean will be within a specified distance of the population mean.
c. A larger sample lowers the population standard deviation.
d. A larger sample has a standard error that is closer to the population standard deviation.

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Answer from: liddopiink1

Using the normal distribution and the central limit theorem, it is found that:

0.2886 = 28.86% probability that a sample size of 20 has a sample mean within $200 of the population mean.

0.4448 = 44.48% probability that a sample size of 50 has a sample mean within $200 of the population mean.

0.5934 = 59.34% probability that a sample size of 100 has a sample mean within $200 of the population mean.

0.9372 = 93.72% probability that a sample size of 500 has a sample mean within $200 of the population mean.

b. A larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean. After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.By the Central Limit Theorem, the standard deviation of the sampling distribution of sample sizes of n is: $s = \frac{\sigma}{\sqrt{n}}$ .

Hence, the probability of sample of size n having a sample mean within 200 is the p-value of $Z = \frac{200}{s}$ subtracted by the p-value of $Z = -\frac{200}{s}$

In this problem:

The standard deviation is $\sigma = 2400$ .

Item a:

For samples of 20, n = 20 , and thus:

$s = \frac{2400}{\sqrt{20}}$

Then

$Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{20}}} = 0.37$

Z = 0.37 has a p-value of 0.6443.Z = -0.37 has a p-value of 0.3557.

0.6443 - 0.3557 = 0.2886.

0.2886 = 28.86% probability that a sample size of 20 has a sample mean within $200 of the population mean.

For samples of 50, n = 50 , and thus:

$s = \frac{2400}{\sqrt{50}}$

Then

$Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{50}}} = 0.59$

Z = 0.59 has a p-value of 0.7224.Z = -0.59 has a p-value of 0.2776.

0.7224 - 0.2776 = 0.4448.

0.4448 = 44.48% probability that a sample size of 50 has a sample mean within $200 of the population mean.

For samples of 100, n = 100 , and thus:

$s = \frac{2400}{\sqrt{100}}$

Then

$Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{100}}} = 0.83$

Z = 0.83 has a p-value of 0.7967.Z = -0.83 has a p-value of 0.2033.

0.7967 - 0.2033 = 0.5934

0.5934 = 59.34% probability that a sample size of 100 has a sample mean within $200 of the population mean.

For samples of 500, n = 500 , and thus:

$s = \frac{2400}{\sqrt{500}}$

Then

$Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{500}}} = 1.86$

Z = 1.86 has a p-value of 0.9686.Z = -1.86 has a p-value of 0.0314.

0.9686 - 0.0314 = 0.9372

0.9372 = 93.72% probability that a sample size of 500 has a sample mean within $200 of the population mean.

Item b:

A larger sample size reduces the standard error, as $s = \frac{\sigma}{\sqrt{n}}$ , that is, the standard error is inversely proportional to the square root of the sample size n, meaning that values are closer to the mean. Thus, the correct option is:

b. A larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

A similar problem is given at link

Answer from: tinyturtles

(a) n : 20 50 100 500

P (-200 < X - μ < 200) : 0.2886 0.4444 0.5954 0.9376

(b) The correct option is (b).

Step-by-step explanation:

Let the random variable X represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.

The mean amount of deductions is, μ = $16,642 and standard deviation is, σ = $2,400.

Assuming that the random variable X follows a normal distribution.

(a)

Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:

For a sample size of n = 20

$P(\mu-200$

=P(-0.37

For a sample size of n = 50

$P(\mu-200$

=P(-0.59

For a sample size of n = 100

$P(\mu-200$

=P(-0.83

For a sample size of n = 500

$P(\mu-200$

=P(-1.86

n : 20 50 100 500

P (-200 < X - μ < 200) : 0.2886 0.4444 0.5954 0.9376

(b)

The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample ( $\bar x$ ) approaches the whole population mean ( $\mu_{x}$ ).

Consider the probabilities computed in part (a).

As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.

So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

Thus, the correct option is (b).

Answer from: Quest

168

step-by-step explanation:

i am pretty sure that the answer is 168

Answer from: Quest

the number of the people going.

step-by-step explanation:

the cost to buy one movie ticket is $7. if the total cost for the movie is a function of how many people go, then the input is : the number of the people going.

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