The 2010 U.S. Census asked every household to report information on each person living there. Suppose for a sample of 30 households selected, the number of persons living in each was reported as follows. 2 3 1 2 6 4 2 1 5 3 2 3 1 2 2 1 3 1 2 2 4 2 1 2 9 3 2 1 1 3 Compute the mean, median, mode, range, lower and upper quartiles, and interquartile range for these data. (Round the mean to 2 decimal places.) Mean

Answers

Answer 1

Answer:

[tex]\bar x =2.53[/tex] -- Mean

[tex]Median= 2[/tex]

[tex]Mode = 2[/tex]

[tex]Range = 8[/tex]

[tex]Q_1 = 1[/tex]

[tex]Q_3 = 3[/tex]

[tex]IQR = 2[/tex]

Step-by-step explanation:

Given

n = 30

Data: 2 3 1 2 6 4 2 1 5 3 2 3 1 2 2 1 3 1 2 2 4 2 1 2 9 3 2 1 1 3

Solving (a): The mean

This is calculated as:

[tex]\bar x =\frac{\sum x}{n}[/tex]

[tex]\bar x =\frac{2 +3 +1 +2 +6 +4 +2 +1 +5 +3 +2 +3+ 1+ 2 +2 +1 +3 +1 +2 +2 +4 +2 +1 +2 +9 +3 +2 +1 +1 +3}{30}[/tex]

[tex]\bar x =\frac{76}{30}[/tex]

[tex]\bar x =2.53[/tex]

Solving (b): The median

First arrange the given data

Arranged: 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 4 4 5 6 9

[tex]Median =\frac{1}{2}(n+1)[/tex]

[tex]Median =\frac{1}{2}(30+1)[/tex]

[tex]Median =\frac{1}{2}(31)[/tex]

[tex]Median = 15.5th[/tex]

This implies that the median is the average of the 15th and 16th item

[tex]Median = \frac{2+2}{2}[/tex]

[tex]Median = \frac{4}{2}[/tex]

[tex]Median= 2[/tex]

Solving (c): The mode

From the given data; 2 has the highest frequency of 11.

So,

[tex]Mode = 2[/tex]

Solving (d): The Range

[tex]Range = Highest\ Data - Lowest\ Data[/tex]

The highest is 9 and the lowest is 1,

So:

[tex]Range = 9 - 1[/tex]

[tex]Range = 8[/tex]

Solving (e): Lower (Q1) and Upper (Q3) quartile.

Q1 is calculated as:

[tex]Q_1 = \frac{1}{4}(n+1)th[/tex]

[tex]Q_1 = \frac{1}{4}(30+1)th[/tex]

[tex]Q_1 = \frac{1}{4}(31)th[/tex]

[tex]Q_1 = 7.75th[/tex]

[tex]Q_1 = 8th\ item[/tex]

[tex]Q_1 = 1[/tex] -- from the arranged data

Q3 is calculated as:

[tex]Q_3 = \frac{3}{4}(n+1)th[/tex]

[tex]Q_3 = \frac{3}{4}(30+1)th[/tex]

[tex]Q_3 = \frac{3}{4}(31)th[/tex]

[tex]Q_3 = 23.25th[/tex]

[tex]Q_3 = 23rd\ item[/tex]

[tex]Q_3 = 3[/tex] -- from the arranged data

Solving (f): The interquartile range (IQR)

[tex]IQR = Q_3 - Q_1[/tex]

[tex]IQR = 3-1[/tex]

[tex]IQR = 2[/tex]


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Answers

Answer:

The slope of the tangent m = -0.0833

Step-by-step explanation:

Step(i):-

Given that the curve

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I hope this is helpful to you

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Answers

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Answers

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Answers

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Step-by-step explanation:

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