Ask question

Ask question

All categories

3 years ago

5

Determine whether or not the given procedure results in a binomial distribution. If not, identify which condition is not met. Su

rveying 2828 people to determine which statistics courses they have taken.

2 answers:

dlinn [17]3 years ago

7 0

Answer:

No, the given procedure doesn't result in a binomial distribution.

Step-by-step explanation:

We can identify if a given procedure results in binomial distribution by observing whether these four conditions are met or not.

Condition 1: Each observation should be independent, it must not depend on the result of previous observation.

Condition 2: The probability of success should stay same from trial to trial.

Condition 3: n is fixed (the number of observations)

Condition 4: There are only two outcomes success or failure.

Surveying 2828 people to determine which statistics courses they have taken.

Condition 3 is satisfied since n is fixed

Condition 1 is satisfied since each observation is independent

Condition 4 is not satisfied since there can be more than 2 answers from the people (statistics courses can be many).

IRINA_888 [86]3 years ago

3 0

Answer:

No, the given procedure doesn't result in binomial distribution.

Step-by-step explanation:

For any procedure to be a binomial distribution, these below four conditions should be met;

<u>Condition 1</u>: Each and every observation should be independent from the other observations.

<u>Condition 2</u>: The probability of success(p) should always be same for each trial.

<u>Condition 3</u>: Number of trials(n) must be fixed.

<u>Condition 4</u>: There must be only two outcomes success or failure.

Now, the procedure given to us is : Surveying 2828 people to determine which statistics courses they have taken.

In this procedure;

Condition 1 is satisfied since each observation is independent from other.

Condition 3 is satisfied as n is fixed for surveying 2828 people.

Condition 4 is not satisfied because there can be more than 2 answers from the people regarding which statistics courses they have taken as there as many statistics courses.

You might be interested in

Divide. Enter the answer in simplest form.<br> 6 2/3 divided by 2 1/2

Roman55 [17]

I can answer but ur not being specific it what is the 6 for and what is the 2 for because there’s a space

8 0

2 years ago

Read 2 more answers

(5x-12)+(3x+30)+(48)

skelet666 [1.2K]

Answer:

8x+66

Step-by-step explanation:

8 0

2 years ago

Neporo4naja [7]

Answer:

a. all matter is made of atoms

b. atoms are indivisible and indestructible

c. atoms of given elements are identical in mass and properties

d. compounds are combination of 2> different types of atoms

2. it's called Planetary model

4 0

3 years ago

Read 2 more answers

PLEASE HELP SOMEONE I DONT KNOW HOW THIS WORKS

Brrunno [24]

Answer:

768

Step-by-step explanation:

7 0

2 years ago

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