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agasfer [191]
2 years ago
6

A random sample of 10 subjects have weights with a standard deviation of kg. What is the variance of their​ weights? be sure to

include the appropriate units with the result.
Mathematics
1 answer:
Vera_Pavlovna [14]2 years ago
7 0

The variance of their​ weights is 128.87270 kg².

It is required to find the variance of their​ weights.

<h3>What is standard deviation ?</h3>

A standard deviation (or σ) is a measure of how dispersed the data is in relation to the mean. Low standard deviation means data are clustered around the mean, and high standard deviation indicates data are more spread out.

Given:

Let n=sample size=10 subjects

Standard deviation=11.352211.3522 kg

Using this formula

Variance=(Standard deviation)²

By the formula we get,

Variance=(11.352211.3522 kg)²

Variance=128.87270 kg²

Therefore, the variance of their​ weights is 128.87270 kg².

Learn more about standard deviation here:

brainly.com/question/16555520

#SPJ4

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9:

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V=A•time

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V=A•t

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Answer:

t=\frac{6.6-7.1}{\frac{1.0}{\sqrt{26}}}=-2.550    

Step-by-step explanation:

Data given and notation  

\bar X=6.6 represent the sample mean

s=1.0 represent the sample standard deviation

n=26 sample size  

\mu_o =7.1 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 7.1 ppm, the system of hypothesis would be:  

Null hypothesis:\mu = 7.1  

Alternative hypothesis:\mu \neq 7.1  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

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Calculate the statistic

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t=\frac{6.6-7.1}{\frac{1.0}{\sqrt{26}}}=-2.550    

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