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Alex777 [14]
2 years ago
14

Bradford Electric Illuminating Company is studying the relationship between kilowatt-hours (thousands) used and the number of ro

oms in a private single-family residence. A random sample of 10 homes yielded the following. Number of Kilowatt-Hours Number of Kilowatt-Hours Rooms (thousands) Rooms (thousands) 12 9 8 6 9 7 10 8 14 10 10 10 6 5 5 4 10 8 7 7 a. Determine the regression equation. B. Determine the number of kilowatt-hours, in thousands, for a six-room house
Mathematics
1 answer:
slamgirl [31]2 years ago
6 0

For the regression equation and e the number of kilowatt-hours, in thousands, for a six-room house  is mathematically given as

Y = 1.33-0.667X and  5.3332

<h3>What is the number of kilowatt-hours, in thousands, for a six-room house?</h3>

Generally, the equation for the  standard deviation  is mathematically given as

sₓ =\sqrt{\frac{\sum\(X-\=X)^2}{n-1} }

= \sqrt{\frac{66.9}{10-1} }

= 2.72

s_y = \sqrt{\frac{36.4}{10-1} }

= 2.01

therefore correlation coff

r = \frac{44.6}{(10-1)(2.72)(2.01)}

r = 0.9038

In conclusion the regression equation is given

b = r_sy/sx

b = (0.9038)(2.01/2.72)

 = 0.667

intercept

a = 7.4-0.667(9.1)

a = 1.33

The regression equation is

Y = 1.33-0.667X

(b) number of kilowatts x = 6 house

Y = 1.33-0.667X

Y = 1.33-0.667(6)

 = 5.3332

Read more about Electric field

brainly.com/question/9383604

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Answer:

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We want to know the probability that X is between 0.64*1850 = 1184 and 0.66*1850 = 1221 (that is, the percentage is between 64 and 66). In order to calculate this, we standarize X so that we can work with a standard normal random variable W ≈ N(0,1). The standarization is obtained by substracting the mean from X and dividing the result by the standard deviation, in other words

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The values of the cummulative function of the standard normal variable W, which we will denote \phi are tabulated and they can be found in the attached file.

Now, we are ready to compute the probability that X is between 1184 and 1221. Remember that, since the standard random variable is symmetric through 0, then \phi(-z) = 1-\phi(z) for each positive value z.

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Therefore, the probability that Polly's Sample will give a result within 1% of the value 65% is 0.6424.

Download pdf
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