The top 15 2018 Congressional race contenders (out of 24) in NJ who raised the most money reported the following amounts of funds raised for their campaigns (in millions): 7.0, 1.2, 6.4, 5.4, 2.1, 4.4, 2.9, 2.3, 1.75, 1.6, 1.5, 1.5, 1.4, 1.1, 0.8 a) What are the Sample Mean, Median, and Mode(s)?b) What is the sample standard Deviation?c) What is the Interval that is 2 Standard Deviations above/below the Mean? (Use 2 decimal places)

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Description: The top 15 2018 Congressional race contenders (out of 24) in NJ who raised the most money reported the following amounts of funds raised for their campaigns (in millions): 7.0, 1.2, 6.4, 5.4, 2.1, 4.4, 2.9, 2.3, 1.75, 1.6, 1.5, 1.5, 1.4, 1.1, 0.8 a)

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A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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The top 15 2018 Congressional race contenders (out of 24) in NJ who raised the most money reported the following amounts of funds raised for their campaigns (in millions):

7.0, 1.2, 6.4, 5.4, 2.1, 4.4, 2.9, 2.3, 1.75, 1.6, 1.5, 1.5, 1.4, 1.1, 0.8

a) What are the Sample Mean, Median, and Mode(s)?
b) What is the sample standard Deviation?
c) What is the Interval that is 2 Standard Deviations above/below the Mean?

(Use 2 decimal places)

Definition Definition Measure of central tendency that is the value that occurs most frequently in a data set. A data set may have more than one mode if multiple categories repeat an equal number of times. For example, in a data set with five item—3, 5, 5, 29, 473—the mode is 5 because it occurs twice and no other value occurs more than once. On a histogram or bar chart, the element with the highest bar represents the mode. Therefore, the mode is sometimes considered the most popular option. The mode is useful for nominal or categorical data (e.g., the most common color car that users purchase), but it is problematic for continuous data because it is more likely not to have any value that is more frequent than the other.

Expert Solution

Step 1

Given data: 7.0, 1.2, 6.4, 5.4, 2.1, 4.4, 2.9, 2.3, 1.75, 1.6, 1.5, 1.5, 1.4, 1.1, 0.8.

Step 2

Sample mean:

$\begin{array}{rcl} x & = & \frac{\sum_{} x}{n} w h e r e n i s s a m p l e s i z e \\ = & \frac{7.0 + . . . + 0.8}{15} \\ = & 2.7567 \end{array}$

Median:

Arrange the observations in an ascending order.

0.8,1.1,1.2,1.4,1.5,1.5,1.6,1.75,2.1,2.3,2.9,4.4,5.4,6.4,7

$\begin{array}{rcl} M e & = & {(\frac{n + 1}{2})}^{t h} v a l u e o f o b s e r v a t i o n \\ = & 8^{t h} v a l u e o f o b s e r v a t i o n \\ = & 1.75 \end{array}$

Mode = 1.5

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