Mean

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ISEE Upper Level Quantitative Reasoning › Mean

Questions 1 - 10

Consider the following set of scores from a physics test. Give the mean of the scores.

$\left \{ 75,73,84,81,68,95,77 \right \}$

Explanation

The mean of the scores can be calculated as:

$\bar{x}=\frac{1}{n}\sum_{i=1}^n{x_{i}$

Where:

$\bar{x}$ is the mean of a data set, $\sum_{}^{}$ indicates the sum of the data values $x_{i}$ , and is the number of data values. So we can write:

$\bar{x}=\frac{1}{n}\sum_{i=1}^n{x_{i}=\frac{75+73+84+81+68+95+77}{7}=\frac{553}{7}=79$

Consider the following set of scores from a physics test. Give the mean of the scores.

$\left \{ 75,73,84,81,68,95,77 \right \}$

Explanation

The mean of the scores can be calculated as:

$\bar{x}=\frac{1}{n}\sum_{i=1}^n{x_{i}$

Where:

$\bar{x}$ is the mean of a data set, $\sum_{}^{}$ indicates the sum of the data values $x_{i}$ , and is the number of data values. So we can write:

$\bar{x}=\frac{1}{n}\sum_{i=1}^n{x_{i}=\frac{75+73+84+81+68+95+77}{7}=\frac{553}{7}=79$

Sally's numeric grade in her economics class is determined by four equally weighted hourly tests, a midterm weighted twice as much as an hourly test, and a final weighted three times as much as an hourly test. The highest score possible on each is 100.

Going into finals week, Sally's hourly test scores are 89, 85, 84, and 87, and her midterm score is 93. What must Sally make on her final at minimum in order to average 90 or more for the term?

It is impossible for Sally to achieve this average this term.

Explanation

Sally's grade is a weighted mean in which her hourly tests have weight 1, her midterm has weight 2, and her final has weight 3. If we call her score on the final, then her course score will be

$\frac{89 + 85 + 84 + 87 + 2 \cdot 93 + 3x}{1 + 1 + 1 + 1 + 2 + 3}$ ,

which simplifies to

$\frac{89 + 85 + 84 + 87 + 186 + 3x}{9} = \frac{3x + 531}{9}$ .

Since Sally wants at least a 90 average for the term, we can set up and solve the inequality:

$\frac{3x + 531}{9} \geq 90$

$\frac{3x + 531}{9} \cdot 9 \geq 90 \cdot 9$

$3x + 531 \geq 810$

$3x + 531-531 \geq 810-531$

$3x \geq 279$

$3x \div 3 \geq 279 \div 3$

$x \geq 93$

Sally must score at least 93 on the final.

Going into finals week, Sally's hourly test scores are 89, 85, 84, and 87, and her midterm score is 93. What must Sally make on her final at minimum in order to average 90 or more for the term?

It is impossible for Sally to achieve this average this term.

Explanation

Sally's grade is a weighted mean in which her hourly tests have weight 1, her midterm has weight 2, and her final has weight 3. If we call her score on the final, then her course score will be

$\frac{89 + 85 + 84 + 87 + 2 \cdot 93 + 3x}{1 + 1 + 1 + 1 + 2 + 3}$ ,

which simplifies to

$\frac{89 + 85 + 84 + 87 + 186 + 3x}{9} = \frac{3x + 531}{9}$ .

Since Sally wants at least a 90 average for the term, we can set up and solve the inequality:

$\frac{3x + 531}{9} \geq 90$

$\frac{3x + 531}{9} \cdot 9 \geq 90 \cdot 9$

$3x + 531 \geq 810$

$3x + 531-531 \geq 810-531$

$3x \geq 279$

$3x \div 3 \geq 279 \div 3$

$x \geq 93$

Sally must score at least 93 on the final.

David's course score in a chemistry course is the mean of five tests. He has scored and . What must he score on the fifth test to be assured of a course score of ?

Explanation

Let be the minimum fifth test score. Then the average of the scores is

$\frac{N + 72 + 81 + 79 + 85 }{5} = 80$ .

$\frac{N + 317 }{5} = 80$

$\frac{N + 317 }{5} \cdot 5 = 80 \cdot 5$

, the minimum score David needs.

David's course score in a chemistry course is the mean of five tests. He has scored and . What must he score on the fifth test to be assured of a course score of ?