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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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Explanation:

Use distributive property:

(5x-2)(4x^2 - 3x-2)
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For a finite sequence of nonzero numbers, the number of variations in sign is defined as the number of pairs of consecutive term
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Answer:

The no. of variations in sign is 3

Step-by-step explanation:

Variation in sign occurs when every single time a negative product is produced by a pair of consecutive terms.

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Therefore, we take the product of the pair of consecutive terms as:

1\times (-3) = - 3, variation in sign

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3 years ago
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Answer:

3.92

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Find the sum of the absolute values of the differences.

Divide the sum of the absolute values of the differences by the number of data values.

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For this case we have the following functions:
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y2 = x ^ 2 + 2

y3 = 2 ^ x
 
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 For y1:
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y1 = 2
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y2 = 0 + 2

y2 = 2
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y2 = 25 + 2

y2 = 27
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y3 = 32
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 Answer:
 
When x = 0, y1 = y2
 
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When x = -1, y3 <y1 <y2
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