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3 years ago

Evaluate the summation of negative 2 n minus 3, from n equals 2 to 10..

1 answer:

5 0

$\bf \sum\limits_{n=2}^{10}~-2n-3$

so, let's first change the bounds from 1 to 9, so we'll be dropping the indices by 1, in order to do the variable "n".

however, let's use a variable hmmm say "k", where "k" is the value of "n" beginning at 1, thus k = n - 1.

however, if k = n - 1, then k + 1 = n. So let's use those fellows then,

$\bf \sum\limits_{n=2}^{10}~-2n-3\qquad 
\begin{cases}
k=n-1\\
k+1=n
\end{cases}\implies \sum\limits_{k=1}^{9}~-2\stackrel{n}{(k+1)}-3
\\\\\\
\sum\limits_{k=1}^{9}~-2k-2-3\implies \sum\limits_{k=1}^{9}~-2k-5\implies \sum\limits_{k=1}^{9}~-2k-\sum\limits_{k=1}^{9}~5
\\\\\\
-2\sum\limits_{k=1}^{9}~k-\sum\limits_{k=1}^{9}~5\implies -2\left( \cfrac{9(9+1)}{2} \right)\qquad -\qquad (9\cdot 5)
\\\\\\
-2(45)-45\implies -90-45\implies -135$

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Usimov [2.4K]

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Step-by-step explanation:

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3 years ago

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3 years ago

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as t

skad [1K]

Answer:

a) 0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

b) 0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

c) 0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

d) None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

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.

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as the population mean and assume the population standard deviation of preparation fees is $100.

This means that $\mu = 273, \sigma = 100$

A) What is the probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 30, s = \frac{100}{\sqrt{30}}$

The probability is the p-value of Z when X = 273 + 16 = 289 subtracted by the p-value of Z when X = 273 - 16 = 257. So

X = 289

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

By the Central Limit Theorem

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

$Z = \frac{289 - 273}{\frac{100}{\sqrt{30}}}$

$Z = 0.88$

$Z = 0.88$ has a p-value of 0.8106

X = 257

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

$Z = \frac{257 - 273}{\frac{100}{\sqrt{30}}}$

$Z = -0.88$

$Z = -0.88$ has a p-value of 0.1894

0.8106 - 0.1894 = 0.6212

0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

B) What is the probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 50, s = \frac{100}{\sqrt{50}}$

X = 289

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

By the Central Limit Theorem

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

$Z = \frac{289 - 273}{\frac{100}{\sqrt{50}}}$

$Z = 1.13$

$Z = 1.13$ has a p-value of 0.8708

X = 257

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

$Z = \frac{257 - 273}{\frac{100}{\sqrt{50}}}$

$Z = -1.13$

$Z = -1.13$ has a p-value of 0.1292

0.8708 - 0.1292 = 0.7416

0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

C) What is the probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 100, s = \frac{100}{\sqrt{100}}$

X = 289

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

By the Central Limit Theorem

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

$Z = \frac{289 - 273}{\frac{100}{\sqrt{100}}}$

$Z = 1.6$

$Z = 1.6$ has a p-value of 0.9452

X = 257

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

$Z = \frac{257 - 273}{\frac{100}{\sqrt{100}}}$

$Z = -1.6$

$Z = -1.6$ has a p-value of 0.0648

0.9452 - 0.0648 =

0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

D) Which, if any of the sample sizes in part (a), (b), and (c) would you recommend to ensure at least a .95 probability that the same mean is withing $16 of the population mean?

None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

6 0

3 years ago

huong and kathyrn each day impeoved their yards by planting rose bushes and ornamental grass. they bought their supplies from th

marta [7]

Answer:

One rose bush = $3

One bunch of ornamental grass = $4

Step-by-step explanation:

We can set up a system of equations to solve for this. Let cost of 1 rose bush be r and cost of 1 ornamental grass be g. Thus we can form 2 equations:

$14r+7g=70\\5r+14g=71$

We can multiply the first equation by -2 so that we can eliminate g. Thus, we have:

$(*-2)14r+7g=70\\5r+14g=71\\-------\\-28r-14g=-140\\5r+14g=71\\--------\\-23r=-69\\r=\frac{-69}{-23}=3$

Thus, 1 rose bush is $3. Now plugging in r = 3 into the first original equation, we can solve for g:

$14r+7g=70\\14(3)+7g=70\\42+7g=70\\7g=70-42\\7g=28\\g=\frac{28}{7}=4$

Thus, 1 bunch of ornamental grass is $4.

5 0

3 years ago