Mathematics: Miguel has an aquarium in the shape of a rectangular prism. the base is 30.25 inches long and 12.5 inches wide. the aquarium is 12.75 inches high. what is the volume of the aquarium to the nearest cubic inches?

Miguel has an aquarium in the shape of a rectangular

jimena15

26.09.2022

$\frac{(21)(5.437)^2}{41.402} \leq \sigma^2 \leq \frac{(21)(5.437)^2}{8.034}$

$14.996 \leq \sigma^2 \leq 77.278$

And the confidence interval for the deviation would be obtained taking the square root of the last result and we got:

3.9<σ<8.8

Step-by-step explanation:

Data given:

65.2 71.9 72.8 73.1 73.1 73.5 75.5 75.7 75.8 76.1 76.2 76.2 77.0 77.9 78.1 79.6 79.7 79.9 80.1 82.2 83.7 93.8

The sample mean would be given by:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

We can calculate the sample deviation with this formula:

$s = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

And we got:

s=5.437 represent the sample standard deviation

$\bar x$ represent the sample mean

n=22 the sample size

Confidence=99% or 0.99

The confidence interval for the population variance is given by:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom given by:

df=n-1=22-1=21

The Confidence is 0.99 or 99%, the value of significance is $\alpha=0.01$ and $\alpha/2 =0.005$ , and the critical values are:

$\chi^2_{\alpha/2}=41.402$

$\chi^2_{1- \alpha/2}=8.034$

And the confidence interval would be:

$\frac{(21)(5.437)^2}{41.402} \leq \sigma^2 \leq \frac{(21)(5.437)^2}{8.034}$

$14.996 \leq \sigma^2 \leq 77.278$

And the confidence interval for the deviation would be obtained taking the square root of the last result and we got:

3.9<σ<8.8

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