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Chapter 3: Problem 38

Use the five test scores of 65,70,71 \(75,\) and 95 to answer the following questions: (a) Find the sample mean. (b) Find the median. (c) Which measure of central tendency better describes the typical test score? (d) Suppose the professor decides to curve the exam by adding 4 points to each test score. Compute the sample mean based on the adjusted scores. (e) Compare the unadjusted test score mean with the curved test score mean. What effect did adding 4 to each score have on the mean?

Short Answer

Expert verified
The sample mean is 75.2. The median is 71. Adding 4 points increased the sample mean to 79.2.

Step by step solution

01

- List the Test Scores

The given test scores are 65, 70, 71, 75, and 95.
02

- Calculate the Sample Mean

Add all the test scores and then divide by the number of scores. \[ \text{Sample Mean} = \frac{65+70+71+75+95}{5} = \frac{376}{5} = 75.2 \]
03

- Find the Median

Order the test scores from smallest to largest: 65, 70, 71, 75, 95. The median is the middle number, which is 71.
04

- Determine the Better Measure of Central Tendency

Compare the mean (75.2) and the median (71). The measure of central tendency that better describes the typical test score is the one that is least affected by outliers. Since 95 is relatively high, the median of 71 may better represent the typical score.
05

- Adjust the Test Scores

Add 4 points to each test score: \[ 65+4 = 69, \; 70+4 = 74, \; 71+4 = 75, \; 75+4 = 79, \; 95+4 = 99 \] The new scores are 69, 74, 75, 79, and 99.
06

- Calculate the Adjusted Sample Mean

Add all the adjusted test scores and then divide by the number of scores. \[ \text{Adjusted Sample Mean} = \frac{69+74+75+79+99}{5} = \frac{396}{5} = 79.2 \]
07

- Compare the Means

Compare the unadjusted mean (75.2) with the adjusted mean (79.2). Adding 4 to each score increased the mean by 4 points.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean
The mean of a set of numbers is often referred to as the arithmetic average. It provides a central value for the data. To find the mean, you sum all the values and then divide by the count of values. In this exercise, we had five test scores, which were 65, 70, 71, 75, and 95. Summing these values yields 376. Dividing 376 by 5 gives a mean of 75.2. This means that the average score across these five tests is 75.2. This calculation is useful because it gives a single number to represent the overall performance within a group of scores. Remember, the mean can sometimes be affected by extremely high or low values, known as outliers.
Median
The median is another type of measure of central tendency. The median is the middle value in a list of numbers that has been ordered from smallest to largest. For an odd number of values, such as the five test scores given (65, 70, 71, 75, 95), the median is the third value, which is 71. If there were an even number of values, you would take the average of the two middle numbers. The median is particularly useful because it is not affected by outliers or extremely high or low values. This makes it a reliable measure of central tendency when the data includes outliers.
Measure of Central Tendency
Measures of central tendency are statistics that describe the center or average of a distribution. The most common measures are the mean, median, and mode. In this exercise, we compared the mean and the median to determine which better represents the typical test score. While the mean was 75.2, the median was 71. The test scores included an outlier (95), which pulled the mean upwards, making it higher than most of the scores. Therefore, in this case, the median better represented the typical test score because it was not influenced by the high value of 95.
Adjusted Scores
Sometimes scores need to be adjusted, such as when a professor decides to add points to everyone's score to curve an exam. In our exercise, 4 points were added to each score, resulting in new scores: 69, 74, 75, 79, and 99. By summing these adjusted scores, we get 396. Dividing by the number of scores (5) gives a new mean of 79.2. Comparing this new mean with the original mean of 75.2, we can see that adding 4 points to each score increased the mean by 4 points as well. This shows that uniform adjustments to all scores affect the mean by the same amount added, making a straightforward way to adjust average performance across a group.

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Most popular questions from this chapter

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