3 1 R7 12 8 27 - 2 4 1 2 O 00 7 4 0 7

Time spent using e-mail per session is normally distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that

liq [111]

Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 11, \sigma = 3$

a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 25, s = \frac{3}{\sqrt{25}} = 0.6$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.6}$

$Z = 0.33$

$Z = 0.33$ has a pvalue of 0.6293.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.6}$

$Z = -0.33$

$Z = -0.33$ has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

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

$Z = \frac{11 - 11}{0.6}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

X = 10.5

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

$Z = \frac{10.5 - 11}{0.6}$

$Z = -0.83$

$Z = -0.83$ has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 100, s = \frac{3}{\sqrt{100}} = 0.3$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.3}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.3}$

$Z = -0.67$

$Z = -0.67$ has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

5 0

3 years ago

The following graphs have a scale assigned to them: The area of each grid

dezoksy [38]

The graphs that are density curves for a continuous random variable are: Graph A, C, D and E.

<h3>How to determine the density curves?</h3>

In Geometry, the area of the density curves for a continuous random variable must always be equal to one (1). Thus, we would test this rule in each of the curves:

Area A = (1 × 5 + 1 × 3 + 1 × 2) × 0.1

Area A = 10 × 0.1

Area A = 1 sq. units (True).

For curve B, we have:

Area B = (3 × 3) × 0.1

Area B = 9 × 0.1

Area B = 0.9 sq. units (False).

For curve C, we have:

Area C = (3 × 4 - 2 × 1) × 0.1

Area C = 10 × 0.1

Area C = 1 sq. units (False).

For curve D, we have:

Area D = (1 × 4 + 1 × 3 + 1 × 2 + 1 × 1) × 0.1

Area D = 10 × 0.1

Area D = 1 sq. units (True).

For curve E, we have:

Area E = (1/2 × 4 × 5) × 0.1

Area E = 10 × 0.1

Area E = 1 sq. units (True).

Read more on density curves here: brainly.com/question/26559908

#SPJ1

6 0

2 years ago

A card is randomly selected from a standard

shepuryov [24]

Answer:

3 / 13

Step-by-step explanation:

Hello!

There are 52 cards in a deck with half being red and half being black. The question ask for when a black card is drawn so we divide the total by 2

52/ 2 = 26

There are 6 black face cards in a deck so to get the probability one will be drawn we put how many there are over the total amount

6/26 we can simplify this

3 / 13

The answer is 3 / 13

Hope this helps!

8 0

3 years ago

Read 2 more answers

What is 5.65 into a mixed number

kodGreya [7K]

5.65 in a mixed number is 5 13/20

7 0

3 years ago

Read 2 more answers

I need help this is the question.

tino4ka555 [31]

Let x = 3.191919…. Then 100x = 319.191919…, and we have

100x - x = 319.191919… - 3.191919…

99x = 316

x = 316/99

Next, we have

316 = 297 + 19 = 3 × 99 + 19

so

316/99 = (3 × 99 + 19)/99 = 3 + 19/99

4 0

3 years ago

3 1 R7 12 8 27 - 2 4 1 2 O 00 7 4 0 7​

3 1 R7 12 8 27 - 2 4 1 2 O 00 7 4 0 7