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GenaCL600 [577]
3 years ago
6

A normal curve with a mean of 500 and a standard deviation of 100 is shown. Shade the region under the curve within one standard

deviation of the mean. How much of the data falls within the shaded region?

Mathematics
1 answer:
N76 [4]3 years ago
6 0

Answer:

And for this case we want to find how much of the data within the shaded region given in the figure attached.

And we can use the zscore formula given by:

z = \frac{X -\mu}{\sigma}

And if we find the z score for the limits we got:

z = \frac{400-500}{100}=-1

z = \frac{600-500}{100}= 1

And we want to find this probability using the normal standard distribution or excel:

P(-1

So we would expect about 68.2 % of the data within one deviation from the mean

Step-by-step explanation:

For this case we know that the variable of interest let's say x follows a normal distribution with the following parameters:

X \sim N(\mu = 500, \sigma =100)

And for this case we want to find how much of the data within the shaded region given in the figure attached.

And we can use the zscore formula given by:

z = \frac{X -\mu}{\sigma}

And if we find the z score for the limits we got:

z = \frac{400-500}{100}=-1

z = \frac{600-500}{100}= 1

And we want to find this probability using the normal standard distribution or excel:

P(-1

So we would expect about 68.2 % of the data within one deviation from the mean

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Can anyone answer this
erik [133]

Answer:

Step-by-step explanation:

1. 10² + 7² = x²

 149 = x²

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2. x² + 19² = 21²

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 x = √80 = 4√5

3. x² + 16² = 27²

 x² = 27² - 16² = 473

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5 0
3 years ago
Quit smoking: In a survey of 444 HIV-positive smokers, 202 reported that they had used a nicotine patch to try to quit smoking.
Flauer [41]

Answer:

The p-value of the test is of 0.0287 < 0.05(standard significance level), which means that it can be concluded that less than half of HIV-positive smokers have used a nicotine patch.

Step-by-step explanation:

Test if less than half of HIV-positive smokers have used a nicotine patch:

At the null hypothesis, we test if the proportion is of at least half, that is:

H_0: p \geq 0.5

At the alternative hypothesis, we test if the proportion is below 0.5, that is:

H_1: p < 0.5

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.5 is tested at the null hypothesis:

This means that \mu = 0.5, \sigma = \sqrt{0.5*(1-0.5)} = 0.5

In a survey of 444 HIV-positive smokers, 202 reported that they had used a nicotine patch to try to quit smoking.

This means that n = 444, X = \frac{202}{444} = 0.455

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.455 - 0.5}{\frac{0.5}{\sqrt{444}}}

z = -1.9

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion below 0.455, which is the p-value of z = -1.9.

Looking at the z-table, z = -1.9 has a p-value of 0.0287.

The p-value is of 0.0287 < 0.05(standard significance level), which means that it can be concluded that less than half of HIV-positive smokers have used a nicotine patch.

5 0
3 years ago
Can you figure this out? And then show your work plz?
Salsk061 [2.6K]

Step-by-step explanation:

You can use PEMDAS to solve this problem from left to right.

First, do 15 ÷ 3 to get 5.

Next, do 2 x 10 to get 20.

Lastly, add 5 and 20 to get 25.

<u>15 ÷ 3</u> + 2 x 10

5 + <u>2 x 10</u>

<u>5 + 20</u>

25

3 0
3 years ago
Read 2 more answers
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