Question

for exercise 7, use the table to answer the question. yuzuki counts the number of raisins in 18 bowls of cereal she eats and records the results in the table. 7. the number of raisins in the 19th bowl of cereal yuzuki eats increases the median number of raisins. what is the least possible number of raisins in her next bowl of cereal? how do you know? number of raisins: 15, 16, 15, 18, 20, 17; 20, 18, 19, 12, 16, 14; 12, 11, 16, 18, 21, 17

Explanation:

Step1: Order the original data

First, we list the number of raisins from the 18 bowls in ascending order: \(11, 12, 12, 14, 15, 15, 16, 16, 16, 17, 17, 18, 18, 18, 19, 20, 20, 21\)

Step2: Find the median of 18 data points

For a set with \(n = 18\) (even number of data points), the median is the average of the \(\frac{n}{2}\)-th and \((\frac{n}{2}+ 1)\)-th values. So, the median is the average of the 9th and 10th values. The 9th value is \(16\) and the 10th value is \(17\), so the median is \(\frac{16 + 17}{2}=\frac{33}{2}=16.5\)

Step3: Consider the 19th data point

When we add a 19th data point, the new set will have \(n = 19\) (odd number of data points), and the median will be the 10th value. We want the new median to be greater than \(16.5\), so the 10th value should be at least \(17\).
Let the number of raisins in the 19th bowl be \(x\). We insert \(x\) into the ordered list. To make the 10th value \(17\), we need to ensure that at least 10 values are less than or equal to the median (after insertion).
Looking at the original ordered list, the first 9 values are \(11, 12, 12, 14, 15, 15, 16, 16, 16\). If we add \(x\), we want the 10th value to be \(17\). So \(x\) should be at least \(17\) (because if \(x = 17\), when we re - order the list with 19 elements, the 10th element will be \(17\)). Let's check:
If \(x = 17\), the new ordered list (19 elements) will be: \(11, 12, 12, 14, 15, 15, 16, 16, 16, 17, 17, 17, 18, 18, 18, 19, 20, 20, 21\). The median (10th value) is \(17\), which is greater than the original median of \(16.5\). If \(x\) is less than \(17\), say \(x = 16\), the new ordered list will be \(11, 12, 12, 14, 15, 15, 16, 16, 16, 16, 17, 17, 18, 18, 18, 19, 20, 20, 21\), and the median (10th value) is \(16\), which is less than \(16.5\), so it does not increase the median.

Answer:

The least possible number of raisins in her next bowl of cereal is \(17\). We know this by first finding the median of the original 18 - data set, then analyzing how adding a 19th data point affects the median (since for 19 data points, the median is the 10th value in the ordered list) and determining the smallest value that makes the new median greater than the original median.