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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

uhm what exactly is the question?