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Yuki888 [10]
3 years ago
14

Pam says she will charge 125 plus 15 an hour to paint a house. Michelle will charge $50 plus $20 per hour to paint a house. Afte

r how many hours will their cost be the same
Mathematics
1 answer:
aniked [119]3 years ago
3 0
Let us assume the number of hours after which the cost of Pam and Michelle will be same = x
Now let us concentrate on the information's already given in the problem.
Fixed charge taken by Pam = $125
Charge per hour taken by Pam = $15
Fixed charge taken by Michelle = $50
Charge per hour taken by Michelle = $20
Then
125 + 15x = 50 + 20x
20x - 15x = 125 - 50
5x = 75
x = 75/5
  = 15
So after 15 hours the charge taken by Pam and Michelle will be the same. I hope the procedure for doing this type of problem is clear to you.
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From the data given, we estimate the population mean and population standard deviation. Then, we use this estimate to find a 95% confidence interval for the population variance and the population standard deviation.

Sample:

Salaries in millions of dollars: 2.2, 1.5, 0.5, 1.3, 2.4, 1.5, 2.7, 0.3, 2.0, 0.3

Question a:

The mean is the sum of all values divided by the number of values. So

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The sample mean salary is of 1.42 million.

Question b:

The standard deviation is the square root of the difference squared between each value and the mean, divided by one less than the number of values.

So

s = \sqrt{\frac{(2.2-1.42)^2 + (1.5-1.42)^2 + (0.5-1.42)^2 + (1.3-1.42)^2 + (2.4-1.42)^2 + (1.5-1.42)^2 + (2.7-1.42)^2 + ...}{9}} = 0.8772

Thus, the estimate for the population standard deviation is of 0.8772 million.

Question c:

The sample size is n = 10

The significance level is \alpha = 1 - 0.05 = 0.95

The estimate, which is the sample standard deviation, is of s = 0.8772.

Now, we have to find the critical values for the Pearson distribution. They are:

\chi^2_{\frac{\alpha}{2},n-1} = \chi^2_{0.025,9} = 19.0228

\chi^2_{1-\frac{\alpha}{2},n-1} = \chi^2_{0.975,9} = 2.7004

The confidence interval for the population variance is:

\frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2},n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2},n-1}}

\frac{9*0.8772^2}{19.0228} < \sigma^2 < \frac{9*0.8772^2}{2.7004}

0.3641 < \sigma^2 < 2.5646

Thus, the 95% confidence interval for the population variance is (0.3641, 2.5646)

Question d:

Standard deviation is the square root of variance, so:

\sqrt{0.3641} = 0.6034

\sqrt{2.5646} = 1.6014

The 95% confidence interval for the population standard deviation is (0.6034, 1.6014).

For more on confidence intervals for population mean/standard deviation, you can check brainly.com/question/13807706

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