Big Ideas Math Integrated I, 2016
BI
Big Ideas Math Integrated I, 2016 View details
Cumulative Assessment

Exercise 7 Page 375

Practice makes perfect
a We will begin by looking at the data set of the scores in a video game.
Video Game Scores
36 28
48 x
42 57
63 52
Given that the scores can be between 0 and 100, we are asked to fill in a value for x to make a mean of 45.5. The mean x is given by the sum of the data values divided by the number of data values. Let's write the expression for the mean and simplify.
x=Sum of Values/Number of Values
x=36+48+42+63+28+x+57+52/8
x=326+x/8
Since we want a mean of 45.5, we can substitute 45.5 for x and solve the equation for x. Let's do it!
x=326+x/8
45.5=326+x/8
Solve for x
364=326+x
38=x
x=38
Therefore, when x=38 the mean of the scores is 45.5.
b In this part, we are asked to fill in a value for x to have a median of 47. The median is the middle value when the data values are in numerical order. If the data set has an even number of values, the median is the mean of the two middle values. Let's order the data values without considering x.
28 36 42 48 52 57 63 Note that if we add x we will have an even number of values. Additionally, by adding it just before 48, we will have the median in terms of x. 28 36 42 x 48 52 57 63 Therefore, the median of the data will be the mean of x and 48. x+ 48/2 Now, since we want a median of 47 we can equate the above expression to 47 and solve it for x.
47=x+48/2
Solve for x
94=x+48
46=x
x=46
Therefore, when x=46 the median of the scores is 47.
c Now, we are asked to fill in a value for x to have a mode of 63. The mode is the most common value or values in a data set. Let's recall our data.

36 48 42 63 28 x 57 52 Note that every value occurs one time, including 63. If we set x=63 the mode of the data will become 63. 36 48 42 63 28 63 57 52 This is because now 63 will occur more often than the other values — that is, twice. Therefore, when x=63 the mode of the scores is 63.

d The range of a data set is the difference of the greatest and least value. If we do not consider x in our data set, the greatest and least values are 63 and 28, respectively. Let's calculate the difference between these values.
63-28=35

As we can see, these values do not give us a range of 71. To fix this, we can set x to be the greatest value of the data. By doing so the range of the data will be the difference between x and 28. x-28 Now, since we want the range to be 71 we can equate the above expression to 71 and solve it for x. 71=x-28 ⇒ x=99 Therefore, when x=99 the range of the scores is 71.