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3 years ago

9. The following data are the ages of eight male

Mathematics

2 answers:

Bess [88]3 years ago

8 0

The mean is 26 and the MAD is 3.75

dybincka [34]3 years ago

5 0

Answer:

The mean is 26 The mean absolute deviation is 3,2,1,5,5,10,3

Step-by-step explanation:

To find the mean you have to add all the numbers together and divide by 8 because there are 8 number so to find the mean absolute deviation of the data, start by finding the mean of the data set.

Find the sum of the data values, and divide the sum by the number of data values.

Find the absolute value of the difference between each data value and the mean: |data value – mean|.

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tankabanditka [31]

Answer:

23.4 cm

Step-by-step explanation:

If the base of the triangle is half of that of a given side, then we need to divide the given side by 2 to get the base.

46.8

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23.4

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It was estimated that 650 people would attend bowling night, but 676 people actually attended.

Cerrena [4.2K]

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2 years ago

The expression 7! is equal to which of the following

Sladkaya [172]

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3 years ago

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Question 3<br> Find 250% of 4.4

alexdok [17]

Step-by-step explanation:

$\to \sf \: \: 250\% \: of \: 4.4 \\ \sf \to \frac{250}{100} \times \frac{44}{100} \\ \sf \to \: \frac{11 \cancel{000}}{10 \cancel{000}} \\ \sf \to \: \frac{11}{10} \\ \sf \to \: 1.1 \: ans.$

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2 years ago

The attached Excel file 2013 NCAA BB Tournament shows the salaries paid to the coaches of 62 of the 68 teams in the 2013 NCAA ba

PSYCHO15rus [73]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

The attached Excel file 2013 NCAA BB Tournament shows the salaries paid to the coaches of 62 of the 68 teams in the 2013 NCAA basketball tournament (not all private schools report their coach's salaries). Consider these 62 salaries to be a sample from the population of salaries of all 346 NCAA Division I basketball coaches.

Question 1. Use the 62 salaries from the TOTAL PAY column to construct a 95% confidence interval for the mean salary of all basketball coaches in NCAA Division I.

xbar = $1,465,752

SD = $1,346,046.2

lower bound of confidence interval ________

upper bound of confidence interval _______

Question 2. Coach Mike Krzyzewski's high salary is an outlier and could be significantly affecting the confidence interval results. Remove Coach Krzyzewski's salary from the data and recalculate the 95% confidence interval using the remaining 61 salaries.

xbar = $1,371,191

SD = $1,130,666.5

lower bound of confidence interval _________

upper bound of confidence interval. ________

Answer:

Question 1:

lower bound of confidence interval = $1,124,027

upper bound of confidence interval = $1,807,477

Question 2:

lower bound of confidence interval = $1,081,512

upper bound of confidence interval = $1,660,870

Step-by-step explanation:

Question 1:

The sample mean salary of 62 couches is

$\bar{x} = 1,465,752$

The standard deviation of mean salary is

$s = 1,346,046.2$

The confidence interval for the mean salary of all basketball coaches is given by

$CI = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } )$

Where $\bar{x}$ is the sample mean, n is the sample size, s is the sample standard deviation and $t_{\alpha/2}$ is the t-score corresponding to a 95% confidence level.

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 62 - 1 = 61

From the t-table at α = 0.025 and DoF = 61

t-score = 1.999

So the required 95% confidence interval is

$CI = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\CI = 1,465,752 \pm 1.999 \cdot (\frac{1,346,046.2}{\sqrt{62} } ) \\\\CI = 1,465,752 \pm 1.999 \cdot (170948.04 ) \\\\CI = 1,465,752 \pm 341,725 \\\\LCI = 1,465,752 - 341,725 = 1,124,027 \\\\UCI = 1,465,752 + 341,725 = 1,807,477\\\\$

Question 2:

After removing the Coach Krzyzewski's salary from the data

The sample mean salary of 61 couches is

$\bar{x} = 1,371,191$

The standard deviation of the mean salary is

$s = 1,130,666.5$

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 61 - 1 = 60

From the t-table at α = 0.025 and DoF = 60

t-score = 2.001

So the required 95% confidence interval is

$CI = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\CI = 1,371,191 \pm 2.001 \cdot (\frac{1,130,666.5}{\sqrt{61} } ) \\\\CI = 1,371,191 \pm 2.001 \cdot (144767 ) \\\\CI = 1,371,191 \pm 289,678.8 \\\\LCI = 1,371,191 - 289,678.8 = 1,081,512 \\\\UCI = 1,371,191 + 289,678.8 = 1,660,870\\\\$

6 0

3 years ago