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3 years ago

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A 7000.00 principal is invested in 2 accounts, 1 earning 3%interest the other 7% interest with the total interest for the year 2

62.00how much is invested in each accpunt

1 answer:

MakcuM [25]3 years ago

7 0

Answer:

The amount in the first account is $5,700

The amount in the second account is $1,300

Step-by-step explanation:

Let the mount invested in the first account be $x$ and the one invested in the second account be $y$ .

Then the total principal is $x+y=7000$ ---->(1)

The interest in the first account is $.03*x$

The interest in the first account is $.07*y$

The total interest for the year is 262.00.

This implies that:

$0.03x+0.07y=262$ ---->(2)

We solve the two equations simultaneously to get:

$x=5,700$ and $y=1,300$

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Read 2 more answers

A random sample of n = 64 observations is drawn from a population with a mean equal to 20 and standard deviation equal to 16. (G

dezoksy [38]

Answer:

a) The mean of a sampling distribution of $\\ \overline{x}$ is $\\ \mu_{\overline{x}} = \mu = 20$ . The standard deviation is $\\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2$ .

b) The standard normal z-score corresponding to a value of $\\ \overline{x} = 16$ is $\\ Z = -2$ .

c) The standard normal z-score corresponding to a value of $\\ \overline{x} = 23$ is $\\ Z = 1.5$ .

d) The probability $\\ P(\overline{x}$ .

e) The probability $\\ P(\overline{x}>23) = 1 - P(Z$ .

f) $\\ P(16 < \overline{x} < 23) = P(-2 < Z < 1.5) = P(Z$ .

Step-by-step explanation:

We are dealing here with the concept of <em>a sampling distribution</em>, that is, the distribution of the sample means $\\ \overline{x}$ .

We know that for this kind of distribution we need, at least, that the sample size must be $\\ n \geq 30$ observations, to establish that:

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

In words, the distribution of the sample means follows, approximately, a <em>normal distribution</em> with mean, $\mu$ , and standard deviation (called <em>standard error</em>), $\\ \frac{\sigma}{\sqrt{n}}$ .

The number of observations is n = 64.

We need also to remember that the random variable Z follows a <em>standard normal distribution</em> with $\\ \mu = 0$ and $\\ \sigma = 1$ .

$\\ Z \sim N(0, 1)$

The variable Z is

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [1]

With all this information, we can solve the questions.

Part a

The mean of a sampling distribution of $\\ \overline{x}$ is the population mean $\\ \mu = 20$ or $\\ \mu_{\overline{x}} = \mu = 20$ .

The standard deviation is the population standard deviation $\\ \sigma = 16$ divided by the root square of n, that is, the number of observations of the sample. Thus, $\\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2$ .

Part b

We are dealing here with a <em>random sample</em>. The z-score for the sampling distribution of $\\ \overline{x}$ is given by [1]. Then

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{16 - 20}{\frac{16}{\sqrt{64}}}$

$\\ Z = \frac{-4}{\frac{16}{8}}$

$\\ Z = \frac{-4}{2}$

$\\ Z = -2$

Then, the <em>standard normal z-score</em> corresponding to a value of $\\ \overline{x} = 16$ is $\\ Z = -2$ .

Part c

We can follow the same procedure as before. Then

$\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{23 - 20}{\frac{16}{\sqrt{64}}}$

$\\ Z = \frac{3}{\frac{16}{8}}$

$\\ Z = \frac{3}{2}$

$\\ Z = 1.5$

As a result, the <em>standard normal z-score</em> corresponding to a value of $\\ \overline{x} = 23$ is $\\ Z = 1.5$ .

Part d

Since we know from [1] that the random variable follows a <em>standard normal distribution</em>, we can consult the <em>cumulative standard normal table</em> for the corresponding $\\ \overline{x}$ already calculated. This table is available in Statistics textbooks and on the Internet. We can also use statistical packages and even spreadsheets or calculators to find this probability.

The corresponding value is Z = -2, that is, it is <em>two standard units</em> <em>below</em> the mean (because of the <em>negative</em> value). Then, consulting the mentioned table, the corresponding cumulative probability for Z = -2 is $\\ P(Z$ .

Therefore, the probability $\\ P(\overline{x}$ .

Part e

We can follow a similar way than the previous step.

$\\ P(\overline{x} > 23) = P(Z > 1.5)$

For $\\ P(Z > 1.5)$ using the <em>cumulative standard normal table</em>, we can find this probability knowing that

$\\ P(Z1.5) = 1$

$\\ P(Z>1.5) = 1 - P(Z$

Thus

$\\ P(Z>1.5) = 1 - 0.9332$

$\\ P(Z>1.5) = 0.0668$

Therefore, the probability $\\ P(\overline{x}>23) = 1 - P(Z$ .

Part f

This probability is $\\ P(\overline{x} > 16)$ and $\\ P(\overline{x} < 23)$ .

For finding this, we need to subtract the cumulative probabilities for $\\ P(\overline{x} < 16)$ and $\\ P(\overline{x} < 23)$

Using the previous <em>standardized values</em> for them, we have from <em>Part d</em>:

$\\ P(\overline{x}$

We know from <em>Part e</em> that

$\\ P(\overline{x} > 23) = P(Z>1.5) = 1 - P(Z$

$\\ P(\overline{x} < 23) = P(Z1.5)$

$\\ P(\overline{x} < 23) = P(Z$

$\\ P(\overline{x} < 23) = P(Z$

Therefore, $\\ P(16 < \overline{x} < 23) = P(-2 < Z < 1.5) = P(Z$ .

5 0

3 years ago