How can we protect space shuttles or astronauts from space radiation in the absence of the atmospheric layer? What is the role of gravity in maintaining the atmospheric layer around the earth ?
Please put your answers with clear explanation.
Thank you ! ​

) A B (ii) A E (iii) C D

(iv) D F (v) F A (vi) D E

(vii) F B

\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 correct answer would be b. amelia, because the particals of gas move really fast (charged particals) and gas also has a lot of room (particals move independently) so the only answer that has both of those is amelia's.