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Ymorist [56]
3 years ago
5

Is the confidence interval affected by the fact that the data appear to be from a population that is not normally​ distributed?

A. ​Yes, because the sample size is not large enough. B. ​No, because the population resembles a normal distribution. C. ​Yes, because the population does not exhibit a normal distribution. D. ​No, because the sample size is large enough.
Mathematics
1 answer:
sp2606 [1]3 years ago
7 0

Answer:

D. No, because the sample size is large enough.

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

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".

If the sample size is higher than 30, on this case the answer would be:

D. No, because the sample size is large enough.

And the reason is given by The Central Limit Theorem since states if the individual distribution is normal then the sampling distribution for the sample mean is also normal.

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

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

If the sample size it's not large enough n<30, on that case the distribution would be not normal.

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Using the rules for significant figures what do you get when you subtract 12.63 from 52.2147
Kipish [7]

Answer:

Step-by-step explanation:

6 0
3 years ago
Use implicit differentiation to find the points where the parabola defined by x2−2xy+y2+4x−8y+20=0 has horizontal and vertical t
Komok [63]

Answer:

The parabola has a horizontal tangent line at the point (2,4)

The parabola has a vertical tangent line at the point (1,5)

Step-by-step explanation:

Ir order to perform the implicit differentiation, you have to differentiate with respect to x. Then, you have to use the conditions for horizontal and vertical tangent lines.

-To obtain horizontal tangent lines, the condition is:

\frac{dy}{dx}=0 (The slope is zero)

--To obtain vertical tangent lines, the condition is:

\frac{dy}{dx}=\frac{1}{0} (The slope is undefined, therefore the denominator is set to zero)

Derivating respect to x:

\frac{d(x^{2}-2xy+y^{2}+4x-8y+20)}{dx} = \frac{d(x^{2})}{dx}-2\frac{d(xy)}{dx}+\frac{d(y^{2})}{dx}+4\frac{dx}{dx}-8\frac{dy}{dx}+\frac{d(20)}{dx}=2x -2(y+x\frac{dy}{dx})+2y\frac{dy}{dx}+4-8\frac{dy}{dx}= 0

Solving for dy/dx:

\frac{dy}{dx}(-2x+2y-8)=-2x+2y-4\\\frac{dy}{dx}=\frac{2y-2x-4}{2y-2x-8}

Applying the first conditon (slope is zero)

\frac{2y-2x-4}{2y-2x-8}=0\\2y-2x-4=0

Solving for y (Adding 2x+4, dividing by 2)

y=x+2 (I)

Replacing (I) in the given equation:

x^{2}-2x(x+2)+(x+2)^{2}+4x-8(x+2)+20=0\\x^{2}-2x^{2}-4x+x^{2} +4x+4+4x-8x-16+20=0\\-4x+8=0\\x=2

Replacing it in (I)

y=(2)+2

y=4

Therefore, the parabola has a horizontal tangent line at the point (2,4)

Applying the second condition (slope is undefined where denominator is zero)

2y-2x-8=0

Adding 2x+8 both sides and dividing by 2:

y=x+4(II)

Replacing (II) in the given equation:

x^{2}-2x(x+4)+(x+4)^{2}+4x-8(x+4)+20=0\\x^{2}-2x^{2}-8x+x^{2}+8x+16+4x-8x-32+20=0\\-4x+4=0\\x=1

Replacing it in (II)

y=1+4

y=5

The parabola has vertical tangent lines at the point (1,5)

4 0
4 years ago
Write an equivalent expression for 5+2+2x+2 <br>please help me ​
patriot [66]

Answer:

2x+9

Step-by-step explanation:

You have to combine like terms. 2x stays the same because there are no other like terms, but 5, 2, and 2 can be added together to make 9

8 0
3 years ago
Need help due tomorrow plese
Sav [38]
The answers to the questions

4 0
3 years ago
3. Find the Area of the circle:
gavmur [86]
The area of the circle is 28.26 sq in hope this helps
5 0
3 years ago
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