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FinnZ [79.3K]
2 years ago
12

What is the estimated margin of error, using standard deviation?

Mathematics
1 answer:
OLEGan [10]2 years ago
3 0
Find the mean ( average):
105 + 104 + 110 + 112 + 114 + 106 + 108 +109 = 868
Mean = 868 / 8 = 108.5

Standard deviation:
108.5 - 105 = 3.5, 3.5^2 = 12.25
108.5 - 104 = 4, 4^2 = 16
110 - 108.5 = 1.5, 1.5^2 = 2.25
112 - 108.5 = 3.5, 3.5^2 = 12.25
114 - 108.5 = 5.5, 5.5^2 = 30.25
108.5 - 106 = 2.5, 2.5^2 = 6.25
108.5 - 108 = 0.5, 0.5^2 = 0.25
109 - 108.5 = 0.5, 0.5^2 = 0.25

12.25 + 16 +2.25 + 12.25 + 30.25 +6.25+0.25+0.25 = 79.75
79.75/8 = 9.96875, SQRT(9.96875) = 3.1573

Standard Deviation = 3.1573

108.8 +/- 1.96 *(3.1573/sqrt(8) = 110.687 and 106.312

110.687 - 108.5 = 2.19
108.5 - 106.312 = 2.19
 

Margin of error = +/- 2.19



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Answer:

864 in²

Step-by-step explanation:

Use the pythagorean theorem to find the missing side.

36²+x²=60² (x is the missing length of the side)

1296 + x² = 3600

x² = 2304

x = 48

Then use the area of a triangle formula: base·height/2

Substituting 36 for the height and 48 for the base, we get:

48·36/2, which is <u>864</u>.

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David wants to buy a new tablet computer. He looks at a sales ad and sees that a computer store is selling a tablet computer for
Jlenok [28]

Answer:

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Step-by-step explanation:

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A population has a mean, μ = 89 and a standard deviation,σ = 24.
Cloud [144]

Answer:

\bar X \sim N(\mu=89, \frac{24}{\sqrt{64}}=3)

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

Let X the random variable of interest for a population. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =89,\sigma =24)

We take a sample of n=64 . That represent the sample size.

The sample mean is defined as:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And if we find the expected value and variance for the sample mean we got:

E(\bar X) = \mu

Var(\bar X) = \frac{\sigma^2}{n}[/tex]

The distribution for the sample mean is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=89, \frac{24}{\sqrt{64}}=3)

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The only possible answer is
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3 years ago
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Solve simultaneous equations 3x-2y=17 +2x-2y=10
Zarrin [17]

Answer:

(7;2)

Step-by-step explanation:

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