ISEE Upper Level Quantitative Reasoning - How to find mean

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.

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.

Consider the data set

$\left { 24,61, 27, 57, 34, 37, 63, 50, N\right }$ .

For what value of does this set have a mean of ?

Explanation

The mean of a nine-element data set is the sum of the data elements divided by ; therefore,

$\frac{ N+ 24 + 61 + 27+ 57+ 34+ 37+ 63+ 50 }{9} = m$

$\frac{ N+ 353}{9} = 50$

$\frac{ N+ 353}{9} \cdot 9 = 50\cdot 9$

What is the mean of the set below?

$(2^{2}, 2^{3}, 2^{4}, 2^{5})$

Explanation

The first step to finding the mean is to convert the set to integers.

$(2^{2}, 2^{3}, 2^{4}, 2^{5})$

This simplifies to:

The mean is determined by adding together the numbers in a set and then dividing by the total number of items in that set. This gives us:

$\frac{4+8+16+32}{4}=\frac{60}{4}=15$

The mean of five numbers is . Give their sum.

Explanation

The mean of five numbers is equal to their sum divided by , or as a formula we can write:

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

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:

$\sum_{i=1}^n{x_{i}=n\bar{x}\Rightarrow \sum_{i=1}^n{x_{i}=5\times 32=160$

Michael received the following scores on his last four French tests: 65, 58, 69, 58.

If his mean test score must be a 70 in the class to pass, what must he score on his fifth test?

Explanation

Remember that the mean is the same as the average.

Let be the score he needs on his fifth test. Since we already know what his average needs to be, we can set up the following equation:

$\frac{65+58+69+58+x}{5}=70$

Solve for .

Michael must score on his next test to pass.

Eric has taken four tests in his English class. His scores on the tests are 94, 87, 95, and 89. The upcoming final exam will be weighted as much as each of the previous four tests. What is the lowest score Eric can get on the final exam and still have a mean score of no less than 90?

Explanation

The first step here, since you're working with the average (or mean), is to determine what is the sum of five test scores that will result in Eric earning at least a mean score of 90.

Start with the fact that he will need to make at least a 90 for his average score. Since you already have the average score, you're working backwards. You need to multiply 90 by 5 for the 5 test scores that will be used to get that average:

$90\times 5=450$

This tells you that the five scores - the four test scores and the final exam - will need to add up to at least 450 in total.

You know the four original test scores, so you can add those up:

So you know the four test scores add up to 365. To get an average of at least 90, the fifth score must bring that sum to at least 450. To find the minimum final exam test score, you subtract 365 from 450: