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Crazy boy [7]
3 years ago
9

Each year, taxpayers are able to contribute money to various charities via their IRS tax forms. The following list contains the

amounts of money (in dollars) donated via IRS tax forms by Each year, taxpayers are able to contribute money taxpayers:
2 , 22 , 27 , 31 , 36 , 51 , 57 , 57 , 60 , 62 , 62 , 62 , 73 , 77 , 83 , 95 , 99 , 104 , 105 , 127 , 153 , 162 , 197

(a) For these data, which measures of central tendency take more than one value? Choose all that apply.

Mean

Median

Mode

None of these measures

(b) Suppose that the measurement 197 (the largest measurement in the data set) were replaced by 246. Which measures of central tendency would be affected by the change? Choose all that apply.

Mean

Median

Mode

None of these measures

(c) Suppose that, starting with the original data set, the largest measurement were removed. Which measures of central tendency would be changed from those of the original data set? Choose all that apply.

Mean

Median

Mode

None of these measures

(d) Which of the following best describes the distribution of the original data? Choose only one.

Negatively skewed

Positively skewed

Roughly symmetrical

Mathematics
2 answers:
user100 [1]3 years ago
6 0

Answer:

(a) None of these measures

(b) Mean

(c) Mean and Median

(d) Roughly Symmetrical

Step-by-step explanation:

(a)

Mean

Total number in the set = 23

Summation of the set = 2+22+27+31+36+51+57+57+60+62+62+62+73+77+83+95+99+104+105+127+153+162+197 = 1804

Mean = Sum of set / total no of set

1804/23 = 78.435

Median is the middle number in the set after it had been arranged from lowest to highest

2 , 22 , 27 , 31 , 36 , 51 , 57 , 57 , 60 , 62 , 62 , 62 , 73 , 77 , 83 , 95 , 99 , 104 , 105 , 127 , 153 , 162 , 197

The Median is 62

Mode the value that appear most

Mode is 62

None of them takes more than one value

(b) If 197 is replaced by 246, the set becomes

2 , 22 , 27 , 31 , 36 , 51 , 57 , 57 , 60 , 62 , 62 , 62 , 73 , 77 , 83 , 95 , 99 , 104 , 105 , 127 , 153 , 162 , 246

The mean becomes

Total number in the set = 23

Summation of the set = 2+22+27+31+36+51+57+57+60+62+62+62+73+77+83+95+99+104+105+127+153+162+246= 1853

Mean = Sum of set / total no of set

1853/23 = 80.565

The Median and Mode remains the same.

(c) When the largest measurements are removed, the number of values in the set reduces and this affects the Mean and the Median. The mode will still remain unchanges since it is a small number and appears the most.

monitta3 years ago
3 0

Answer:

A. median and mode

B. Mean

C. Mean

d. Positively skewed

Step-by-step explanation:

a. Mean = sum of all values / numbers of values = 1804 / 23 = 78.4. this eliminates the mean and leaves the median and mode, both of which are 62.  

Median = middlemost value = 62

Mode = most frequent value = 62

b. Only the Mean would be affected. It would move from 78.4 to 80.6. (calculated as 1853 / 23). The median and mode would not be affected as the 62 would remain the middlemost value and the most frequent.

 

c. Removing the largest value would result in the mean changing. The mean would now be 73 (calculated as 1607 / 22). Since the values are no longer 23, but now 22, the median would be the average of the 2 middlemost values, ((62+62) / 2) which would still result in 62 as the median.  

d. we use the Five number summary:  

Median / Interquartile Range = 62

First Quartile / Q1 = (51 +57) / 2 = 54

Third Quartile / Q3 =  (99 + 104)  / 2 = 101.5

Minimum = 2

Maximum = 197 ; therefore Range = (197 - 2) = 195.

Please see attached box and whisker plot. The distribution is positively skewed.

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Answer:

(a) See below

(b) r = 0.9879  

(c) y = -12.629 + 0.0654x

(d) See below

(e) No.

Step-by-step explanation:

(a) Plot the data

I used Excel to plot your data and got the graph in Fig 1 below.

(b) Correlation coefficient

One formula for the correlation coefficient is  

r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}

The calculation is not difficult, but it is tedious.

(i) Calculate the intermediate numbers

We can display them in a table.

<u>    x   </u>    <u>      y     </u>   <u>       xy     </u>    <u>              x²    </u>   <u>       y²    </u>

   36       0.22              7.92               1296           0.05

   67        0.62            42.21              4489           0.40

   93         1.00            93.00           20164           3.46

 433        11.8          5699.4          233289        139.24

 887      29.3         25989.1          786769       858.49

1785      82.0        146370          3186225      6724

2797     163.0         455911         7823209    26569

<u>3675 </u>  <u> 248.0  </u>    <u>   911400      </u>  <u>13505625</u>   <u> 61504        </u>

9965   537.81     1545776.75  25569715   95799.63

(ii) Calculate the correlation coefficient

r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}\\\\= \dfrac{9\times 1545776.75 - 9965\times 537.81}{\sqrt{[9\times 25569715 -9965^{2}][9\times 95799.63 - 537.81^{2}]}} \approx \mathbf{0.9879}

(c) Regression line

The equation for the regression line is

y = a + bx where

a = \dfrac{\sum y \sum x^{2} - \sum x \sum xy}{n\sum x^{2}- \left (\sum x\right )^{2}}\\\\= \dfrac{537.81\times 25569715 - 9965 \times 1545776.75}{9\times 25569715 - 9965^{2}} \approx \mathbf{-12.629}\\\\b = \dfrac{n \sum xy  - \sum x \sum y}{n\sum x^{2}- \left (\sum x\right )^{2}} -  \dfrac{9\times 1545776.75  - 9965 \times 537.81}{9\times 25569715 - 9965^{2}} \approx\mathbf{0.0654}\\\\\\\text{The equation for the regression line is $\large \boxed{\mathbf{y = -12.629 + 0.0654x}}$}

(d) Residuals

Insert the values of x into the regression equation to get the estimated values of y.

Then take the difference between the actual and estimated values to get the residuals.

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    36        0.22        -10                 10

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   142        1.86           -3                  5

  433       11.8             19               -  7

  887     29.3             45               -16  

 1785     82.0            104              -22

2797    163.0            170               -  7

3675   248.0            228               20

(e) Suitability of regression line

A linear model would have the residuals scattered randomly above and below a horizontal line.

Instead, they appear to lie along a parabola (Fig. 2).

This suggests that linear regression is not a good model for the data.

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irina [24]

Using two unit multiplier 628 km is equal to 62800000 cm

<u>Solution:</u>

628 kilometer to centimeter

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