Chapter 1: Q 80. (page 70)

Cowboys the $2009$ roster of the Dallas Cowboys football team included $7$ defensive linemen. Their weights (in pounds) were $306, 305, 315, 303, 318, 309, a n d 285$
Calculate the mean. Show your work. Interpret your result in context.

Short Answer

Expert verified

The mean is $305.86$

Step by step solution

Given information

The weights (in pounds) were $306, 305, 315, 303, 318, 309,$ and $285$

Concept

Formula used= $= \frac{Σ X i}{N}$

Calculation

The following is the average of Joe's first $14$ quiz grades in a marking period:

Mean= $X$

$= \frac{Σ X i}{N} = \frac{x_{1} + x_{2} + x_{3} + . . . . . . . + x_{n}}{N}$

= \frac{306 + 305 + 315 + 303 + 318 + 309 + 285}{7} = \frac{s u m o f o b s e r v a t i o n s}{n} = \frac{2141}{7}

Mean= $305.86$

The mean is a "fair share" or a balance point that shows us how big each observation in the data set would be if the total was distributed evenly among all the observations. There are three observations below the mean and four observations above the mean, as can be seen. If the smallest $(285)$ observation is removed, the mean becomes $309.33$ , which is the data's equilibrium point. The value of mean is reduced by $3.47$ points as a result of one observation. With a mean of $309.33$ for six observations (after excluding the extreme observation), three observations are above $309.33$ and three are below it. As a result, we conclude that the existence of extreme values affects the mean. Outliers are sensitive to the mean.