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Scorpion4ik [409]
1 year ago
8

The lengths of text messages are normally distributed with an unknown population mean. A random sample of text messages is taken

and results in a 95% confidence interval of (23,47) characters. What is the correct interpretation of the 95% confidence interval?
Mathematics
2 answers:
anygoal [31]1 year ago
8 0

Answer:

95% of the text messages have length between 23 units and 47 units.

Step-by-step explanation:

We are given the following in the question:

The lengths of text messages are normally distributed.

95% confidence interval:

(23,47)

Thus, we could interpret the confidence interval as:

About 95% of the text messages have length between 23 units and 47 units.

By Empirical rule for a normally distributed data, about 95% of data lies within 2 standard deviations of mean , thus we can write:

\mu - 2\sigma = 23\\\mu +2\sigma = 47\\\Rightarrow \mu = 35\\\Rightarrow \sigma = 6

Thus, the mean length of text messages is 23 units and standard deviation is 6 units.

Eva8 [605]1 year ago
3 0

Answer:

Correct interpretation is provided below.

Step-by-step explanation:

We are given that the lengths of text messages are normally distributed with an unknown population mean.

Also, a random sample of text messages is taken and results in a 95% confidence interval of (23,47) characters.

Now, the correct interpretation of the 95% confidence interval is that we are 95% confident that the population mean will lie between the 23 and 47 as the confidence interval is estimated taking into consideration the population mean only.

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Answer:

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Step-by-step explanation:

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Therefore:

\dfrac{dA}{dt}=3-\dfrac{4A(t)}{100+2t}\\\\\dfrac{dA}{dt}=3-\dfrac{4A(t)}{2(50+t)}\\\\\dfrac{dA}{dt}=3-\dfrac{2A(t)}{50+t}\\\\\dfrac{dA}{dt}+\dfrac{2A(t)}{50+t}=3

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e^{\int \frac{2}{50+t}dt}=e^{2\ln|50+t|}=e^{\ln(50+t)^2}=(50+t)^2

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(50+t)^2\dfrac{dA}{dt}+(50+t)^2\dfrac{2A(t)}{50+t}=3(50+t)^2\\\{(50+t)^2A(t)\}'=3(50+t)^2\\$Taking the integral of both sides\\\int \{(50+t)^2A(t)\}'= \int 3(50+t)^2\\(50+t)^2A(t)=(50+t)^3+C $ (C a constant of integration)\\Therefore:\\A(t)=(50+t)+C(50+t)^{-2}

Initially, 20 pounds of salt was dissolved in the tank, therefore: A(0)=20

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A(t)=(50+t)-75000(50+t)^{-2}

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A(15)=(50+15)-75000(50+15)^{-2}\\=47.25$ pounds

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Answer:

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Step-by-step explanation:

It is given that:

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