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2 years ago

12 = 2 + d; d = i am so confused pls help

Mathematics

1 answer:

BaLLatris [955]2 years ago

7 0

If your asking what d equals, then it’s 10

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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 a sampling distribution, 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 normal distribution with mean, $\mu$ , and standard deviation (called standard error), $\\ \frac{\sigma}{\sqrt{n}}$ .

The number of observations is n = 64.

We need also to remember that the random variable Z follows a standard normal distribution 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 random sample. 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 standard normal z-score 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 standard normal z-score 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 standard normal distribution, we can consult the cumulative standard normal table 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 two standard units below the mean (because of the negative 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 cumulative standard normal table, 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 standardized values for them, we have from Part d:

$\\ P(\overline{x}$

We know from Part e that

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

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

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

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

5 0

2 years ago

Will mark BRAINLIST only if correct answer

Softa [21]

Answer:

70.60

Step-by-step explanation:

4 0

2 years ago

Which expressions are equivalent to 3x + 2(x − 1) − 4? A.) 3x + 2x - 2 - 4 B.) 5x - 6 C.) 3(x - 4) D.) 3(x + 2) - 6 E.) 6(x - 1)

sasho [114]

Multiplication takes precedence
Execute the command multiplication bracket

3x + 2* (x - 1) - 4 =
3x + 2* x + 2* (-1 ) - 4 =
3x + 2x - 2 - 4 =

We reduce similar words

3x + 2x - 2 - 4 =
5x - 6

Answer B

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2 years ago

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belka [17]

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Triangle HIJ is dilated by a scale factor of 55 to form triangle H'I'J'. Side I'J' measures 7575. What is the measure of side IJ

kvasek [131]

Answer:

If triangle HIJ is dilated about the center of the triangle to create triangle H'I'J', dilated line A'B' will be parallel to AB.Option (D) is correct.

Given:

The options are as follows,

(A). be perpendicular to AB.

(B). lie on the same line as AB.

(C). shift four units to the left.

(D). be parallel to AB.

Step-by-step explanation:

If we dilate triangle HIJ then the dilated triangle is bigger than the triangle HIJ.

The triangle HIJ is similar to triangle to each other.

So the line is parallel to side AB.

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2 years ago

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