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Ugo [173]
2 years ago
14

If population function x has mean M(x)=2 and M(x²)=8, find its standard deviation​

Mathematics
1 answer:
sasho [114]2 years ago
6 0

The standard deviation of the population function is 2

<h3>How to determine the standard deviation?</h3>

The given parameters are:

M(x) = 2

M(x²)=8

The standard deviation is calculated using:

\sigma = \sqrt{M(x^2) - (M(x))^2

Substitute known values

\sigma = \sqrt{8 - 2^2

Evaluate the expression

\sigma = 2

Hence, the standard deviation of the population function is 2

Read more about standard deviation at:

brainly.com/question/15858152

#SPJ1

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3/4 * 5/7 =  15/28  or 0.5357
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Tina wants to save money for school. Tina invests $1,100 in an account that pays an interest rate of 7.25%. How many years will
svlad2 [7]
The amount of Tina money can be expressed in an exponent function like this:
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4 years ago
How do I solve this word problem?
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3 years ago
Simplify and solve each equation for the x
Olegator [25]

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I hope this helps!

8 0
3 years ago
26. Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics
hjlf

The test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

<h3>What are null hypotheses and alternative hypotheses?</h3>

In null hypotheses, there is no relationship between the two phenomenons under the assumption or it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.

Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics course.

Then they are given a post-test once the professor has concluded lecturing on the material.

Pre-test and post-test scores for 4 students in an elementary statistics class are given below.

Then we have

\mu _d = \mu _{post} - \mu _{pre}

Then the null hypotheses and alternative hypotheses will be

H₀: \mu _d = 0

Hₐ: \mu _d > 0

Then the test statistic will be

\rm \overline{x} _d = \dfrac{\Sigma x_d}{n} = \dfrac{15+12+10+1}{4}\\\\\overline{x} _d = 9.5

Then

\rm S_d = 6.02

The test statistic value is given by

\rm t = \dfrac{\overline{x} _d }{\dfrac{S_d}{\sqrtn}} \\\\t = \dfrac{9.5}{\dfrac{6.02}{\sqrt4}}\\\\t = 3.16

Since this is a right-tailed test, so the critical value is given by

\rm t_{n-1}(\alpha ) = t_3 (0.05) = 2.353

Since the test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

Hence, we can conclude that \mu _d > 0 that is test scores have improved.

More about the null hypotheses and alternative hypotheses link is given below.

brainly.com/question/9504281

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7 0
2 years ago
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