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Zigmanuir [339]
1 year ago
5

According to chebyshev's theorem, at least what percent of the observations lie within plus and minus 1.75 standard deviations o

f the mean?

Mathematics
1 answer:
Alborosie1 year ago
4 0
Chebyshev's theorem in statistics states that for many probability distributions, no more than 1/k² of measured values will be k standard deviations away from the mean.

Because the area under the probability distribution curve is equal to 1, Chebyshev's theorem means that the shaded area shown in the figure is equal to 1 - 1/k².

When k = 1.75, the shaded area is
1 - 1/1.75² = 0.7635 = 67.35%

Therefore the percent of the area within +/- 1.75 standard deviations from the mean is
67.35/2 = 33.7%, which is at least 33% of the observations.

Answer:
According to the Chebyshev theorem, at least 33% of the observations lie within +/- standard deviations from the mean.

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An investor believes that investing in domestic and international stocks will give a difference in the mean rate of return. They
arlik [135]

Answer:

So on this case the 90% confidence interval would be given by -4.137 \leq \mu_1 -\mu_2 \leq 2.087  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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".  

\bar X_1 =2.0233 represent the sample mean 1  

\bar X_2 =3.048 represent the sample mean 2  

n1=15 represent the sample 1 size  

n2=15 represent the sample 2 size  

s_1 =4.893387 population sample deviation for sample 1  

s_2 =5.12399 population sample deviation for sample 2  

\mu_1 -\mu_2 parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%

We want to test:

H0: \mu_1 = \mu_2

H1: \mu_1 \neq \mu_2

And we can do this using the confidence interval for the difference of means.

Solution to the problem  

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:  

\bar X_1 -\bar X_2 =2.0233-3.048=-1.0247

The degrees of freedom are given by:

df = n_1 +n_2 -2 = 15+15-2=28  

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,28)".And we see that t_{\alpha/2}=\pm 1.701  

Now we have everything in order to replace into formula (1):  

-1.0247-1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=-4.137  

-1.0247+1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=2.087  

So on this case the 90% confidence interval would be given by -4.137 \leq \mu_1 -\mu_2 \leq 2.087  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

4 0
1 year ago
What is a3 in an arithmetic sequence in which a10=41 and a15=61
USPshnik [31]
\bf \begin{array}{llll}
term&value\\
-----&-----\\
a_{10}&41\\
a_{11}&41+d\\
a_{12}&(41+d)+d\\
&41+2d\\
a_{13}&(41+2d)+d\\
&41+3d\\
a_{14}&(41+3d)+d\\
&41+4d\\
a_{15}&(41+4d)+d\\
&41+5d=61
\end{array}
\\\\\\
41+5d=61\implies 5d=20\implies d=\cfrac{20}{5}\implies \boxed{d=4}\\\\
-------------------------------\\\\

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
d=4\\
n=10\\
a_{10}=41
\end{cases}
\\\\\\
41=a_1+(10-1)4\implies 41=a_1+36\implies \boxed{5=a_1}

thus

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
d=4\\
n=3\\
a_{1}=5
\end{cases}
\\\\\\
a_3=a_1+(3-1)4\implies a_3=5+(3-1)4

and surely you know how much that is.
8 0
2 years ago
Read 2 more answers
Rita and Tina each make $11 an hour working as cashiers at a supermarket. Last week, Rita worked r hours while Tina worked t hou
Sonbull [250]
Rita = (11r + 32) 
 Tina  = 11t
5 0
2 years ago
We are told that the data is representative of the two populations (U.S. males aged 20-29 years and U.S. males aged 75+ years),
tiny-mole [99]

Answer:

The data provide strong evidence that young men weigh more on average than old men in the U.S

Step-by-step explanation:

Given :

The null hypothesis ; H0 : μ1 = μ2

The alternative hypothesis ; H1 : μ1 > μ2

T score = 5.3 ; Pvalue = < 0.0001

The decision region :

If Pvalue < α ; We reject the Null

If Pvalue > α ; We fail to reject the Null

When the α - level isn't stated, we usually assume a α - level of 5%

However, even at lower alpha level of 1% = 0.01 ;

The Pvalue < α

Hence, we can conclude that there is significant evidence that there is difference in the mean weight of young men and old men in the U.S

3 0
1 year ago
The probability that Evan purchases a comic book from the bookstore (event A) is 0.67, and the probability that Glen purchases a
Sveta_85 [38]
The question is asking to choose among the following choices that states the statement that is true about the said information and base on the given data, I would say the answer among them is <span>Events A and B are dependent because P(A and B) do not equal P(A) x P(B). I hope you are satisfied with my answer and feel free to ask for more</span>
7 0
2 years ago
Read 2 more answers
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