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3 years ago

12

A bus picks up passengers at a bus stop every 12 minutes in the morning. Suppose Adrian arrives at the bus stop at a random time

. What is the
probability that he will have to wait 5 minutes or less for the next bus?

These are the options
A. 7/12
B. 5/12
C. 1/5
D. 5/24

I need it right now please

1 answer:

allsm [11]3 years ago

5 0

I’m not too sure but i thinks it’s B

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3 years ago

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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 <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

What's the unit rate for 15 plates in 3 stacks?

andreev551 [17]

Answer:

My pretty sure its 5

(don’t count on it too much, this is my first answer)

Step-by-step explanation:

7 0

3 years ago

The data show the traveler spend- ing in billions of dollars for a recent year for a sample of the states. Find the range, varia

Svetllana [295]

Solution :

Given data :

20.1 33.5 21.7 58.4 23.2 110.8 30.9

24.0 74.8 60.0

n = 10

Range : Arranging from lowest to highest.

20.1, 21.7, 23.2, 24.0, 30.9, 33.5, 58.4, 60.0, 74.8, 110.8

Range = low highest value - lowest value

= 110.8 - 20.1

= 90.7

Mean = $\frac{\sum x}{n}$

$=\frac{20.1+21.7+23.2+24.0+30.9+33.5+58.4+60.0+74.8+110.8}{10}$

$=\frac{457.4}{10}$

$=45.74$

Sample standard deviation :

$S=\sqrt{\frac{1}{n-1}\sum(x-\mu)^2}$

$S=\sqrt{\frac{1}{10-1}(20.1-45.74)^2+(21.7-45.74)^2+(23.2-45.74)^2+(24.0-45.74)^2+(30.9-45)^2}$

$\sqrt{(33.5-45.74)^2+(58.4-45.74)^2+(60.0-45.74)^2+(74.8-45.74)^2+(110.8-45.74)^2}$

$S=\sqrt{\frac{1}{9}(657.4+577.9+508.0+472.6+220.2+149.8+160.2+203.3+844.4+4232.8)}$ $S=\sqrt{\frac{1}{9}(8026.96)}$

$S=\sqrt{891.88}$

S = 29.8644

Variance = $S^2$

$=(29.8644)^2$

= 891.8823

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3 years ago