Mathematics, 19.05.2021 18:10 haileybug0305

An important aspect of a federal economic plan was that consumers would save a substantial portion of the money that they received from an income tax reduction. Suppose that early estimates of the portion of total tax saved, based on a random sampling of 35 economists, had mean 26% and standard deviation 12%. Required:
What is the approximate probability that a sample mean estimate, based on a random sample of n = 35 economists, will lie within 1% of the mean of the population of the estimates of all economists?

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0.3758 = 37.58% probability that a sample mean estimate will lie within 1% of the mean of the population of the estimates of all economists.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean 26% and standard deviation 12%.

This means that $\mu = 26, \sigma = 12$

Sample of 35:

This means that $n = 35, s = \frac{12}{\sqrt{35}}$

What is the approximate probability that a sample mean estimate, based on a random sample of n = 35 economists, will lie within 1% of the mean of the population of the estimates of all economists?

This is the p-value of Z when X = 26 + 1 = 27 subtracted by the p-value of Z when X = 26 - 1 = 25. So

X = 27

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

By the Central Limit Theorem

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

$Z = \frac{27 - 26}{\frac{12}{\sqrt{35}}}$

Z = 0.49

Z = 0.49 has a p-value of 0.6879

X = 25

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

$Z = \frac{25 - 26}{\frac{12}{\sqrt{35}}}$

Z = -0.49

Z = -0.49 has a p-value of 0.3121

0.6879 - 0.3121 = 0.3758

0.3758 = 37.58% probability that a sample mean estimate will lie within 1% of the mean of the population of the estimates of all economists.

Answer from: Quest

the answer is (4,3)

since the rule for r0, 90º is (-y,x) and it was already applied to form the image, all you have to do is reverse the answer using this formula.

Arectangle is transformed according to the rule r0,90 the image of the rectangle has vertices locate

Answer from: Quest

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

hope to and hope that my answer is brainliest.(follow me )

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