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\bar x = 260.1615

x

ˉ

=260.1615

\sigma = 70.69σ=70.69

The confidence interval of standard deviation is: 53.7653.76 to 103.25103.25

Step-by-step explanation:

Given

n =20n=20

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

\bar x = \frac{\sum x}{n}

x

ˉ

=

n

∑x

So, we have:

\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 ++250.61}{20}

x

ˉ

=

20

242.87+212.00+260.93+284.08+194.19+139.16+260.76+436.72+355.36++250.61

\bar x = \frac{5203.23}{20}

x

ˉ

=

20

5203.23

\bar x = 260.1615

x

ˉ

=260.1615

\bar x = 260.16

x

ˉ

=260.16

Solving (b): The standard deviation

This is calculated as:

\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}σ=

n−1

∑(x−

x

ˉ

)

2

\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2++(250.61- 260.1615)^2}{20 - 1}}σ=

20−1

(242.87−260.1615)

2

+(212.00−260.1615)

2

+(260.93−260.1615)

2

+(284.08−260.1615)

2

++(250.61−260.1615)

2

\sigma = \sqrt{\frac{94938.80}{19}}σ=

19

94938.80

\sigma = \sqrt{4996.78}σ=

4996.78

\sigma = 70.69σ=70.69 --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

c =0.95c=0.95

So:

\alpha = 1 -cα=1−c

\alpha = 1 -0.95α=1−0.95

\alpha = 0.05α=0.05

Calculate the degree of freedom (df)

df = n -1df=n−1

df = 20 -1df=20−1

df = 19df=19

Determine the critical value at row df = 19df=19 and columns \frac{\alpha}{2}

2

α

and 1 -\frac{\alpha}{2}1−

2

α

So, we have:

X^2_{0.025} = 32.852X

0.025

2

=32.852 at \frac{\alpha}{2}

2

α

X^2_{0.975} = 8.907X

0.975

2

=8.907 --- at 1 -\frac{\alpha}{2}1−

2

α

So, the confidence interval of the standard deviation is:

\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} } to \sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }

70.69 * \sqrt{\frac{20 - 1}{32.852} to 70.69 * \sqrt{\frac{20 - 1}{8.907}

70.69 * \sqrt{\frac{19}{32.852} to 70.69 * \sqrt{\frac{19}{8.907}

53.7653.76 to 103.25103.25

the answer to your question is,

puerto rico

-mabel < 3

when the outside temperature is too cold, an endothermic animal can generate heat by shivering. endothermic animals must eat much more often than ectothermic animals since it takes energy to maintain a constant body temperature.