Question Video: Calculating Upper and Lower Quartiles for a Set of Data Mathematics

David’s history test scores are 74, 96, 85, 90, 71, and 98. Determine the upper and lower quartiles of his scores.

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Video Transcript

David’s history test scores are 74, 96, 85, 90, 71, and 98. Determine the upper and lower quartiles of his scores.

In order to calculate the upper and lower quartiles for a data set, we firstly need to sort the data into ascending order. In this case, the lowest score was 71. The next lowest score was 74. The remainder of David’s scores in ascending order were 85, 90, 96, and 98. We have six test scores in total, and we know that the median is the middle value.

One way to calculate the median with a small data set is to cross off numbers from either end. We cross off the smallest number and the largest number. We then cross off 74 and 96. This means we’re left with two middle numbers, 85 and 90. The median will be the midpoint of these two numbers. We could work this out on a number line. Alternatively, we can find the average or midpoint of two numbers by finding their sum and dividing by two. This is equal to 87.5. The median of David’s test scores is 87.5.

An alternative way of finding the median, which is useful if we have a large data set, is by using the formula 𝑛 plus one divided by two. This gives us the median position on the list. As there were six values in this question, 𝑛 is equal to six. Six plus one is equal to seven, and dividing by two gives us 3.5. This means that the median will be halfway between the third and fourth value. This confirms that our answer of 87.5 was correct.

As we had six values in total, there are three values less than the median and three values greater than the median. We know that the lower quartile is the center of the bottom half of our data set. As there are three values here, the lower quartile, or Q one, will be the middle one. This is equal to 74. The upper quartile will be the center of the top half of our data set. Once again, we have three numbers above the median. The center number will be the middle one. This is equal to 96.

We can therefore conclude that the upper quartile of David’s history scores was 96 and the lower quartile was 74. Before moving on from this question, let’s consider how we could find the lower quartile and upper quartile position. The position of the lower quartile can be calculated using the formula 𝑛 plus one divided by four or a quarter of 𝑛 plus one. Seven divided by four is equal to 1.75. As this is more than halfway between one and two, we round up to two. The lower quartile will be the second value in our list.

We can calculate the position of the upper quartile using a similar method. Three-quarters of 𝑛 plus one, or three multiplied by 𝑛 plus one divided by four. This is equal to 5.25, which we notice is three times 1.75. As this is less than halfway between five and six, we round down to five. The fifth number in our list will be the upper quartile. This method is particularly useful if we have a large data set.

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