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3 years ago

12

I need help please, how do U do this?

1 answer:

Korolek [52]3 years ago

4 0

Answer:

1) point slope form: y-y1=m(x-x1) so y-3=5/3(x-1)

y-3=5/3x-5/3

y=5/3x+4/3 <---bc 9/3-5/3=4/3

2)Plug in 1,2,3,4,5 for x and solve

y=5/3*1-4/3

y=1/3 (1, 1/3)

y=5/3*2-4/3

y=2 (2,2)

y=5/3*3-4/3

y=11/3 (3, 11/3)

y=5/3*4-4/3

y=16/3 (4, 16,3)

y=5/3*5-4/3

y=24/3 or 8 (5,8)

3) Plot the ordered pairs I wrote above

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The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room

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Given

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See attachment for the formatted data

Solving (a): The mean

This is calculated as:

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

So, we have:

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

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

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Solving (b): The standard deviation

This is calculated as:

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

$\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}$ $\sigma = \sqrt{\frac{94938.80}{19}}$

$\sigma = \sqrt{4996.78}$

$\sigma = 70.69$ --- approximated

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

We have:

$c =0.95$

So:

$\alpha = 1 -c$

$\alpha = 1 -0.95$

$\alpha = 0.05$

Calculate the degree of freedom (df)

$df = n -1$

$df = 20 -1$

$df = 19$

Determine the critical value at row $df = 19$ and columns $\frac{\alpha}{2}$ and $1 -\frac{\alpha}{2}$

So, we have:

$X^2_{0.025} = 32.852$ ---- at $\frac{\alpha}{2}$

$X^2_{0.975} = 8.907$ --- at $1 -\frac{\alpha}{2}$

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 $70.69 * \sqrt{\frac{19}{8.907}$

$53.76$ to $103.25$

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