Question

suppose the 2006 prices, per ton, of wine grapes for different kinds of wine are given in the following table.
variety\tprice
cabernet sauvignon\t$1,100
merlot\t$1,950
zinfandel\t$1,300
pinot noir\t$1,350
chardonnay\t$1,650
sauvignon blanc\t$1,350
what is the median?
$

Explanation:

Step1: Order the prices

First, we list the prices in ascending order: $1,100, $1,300, $1,350, $1,350, $1,650, $1,950.

Step2: Find the median position

There are 6 data points. The median for an even number of data points is the average of the \(\frac{n}{2}\)-th and \((\frac{n}{2}+1)\)-th values, where \(n = 6\). So \(\frac{6}{2}=3\) and \(\frac{6}{2}+1 = 4\).

Step3: Calculate the median

The 3rd value is $1,350 and the 4th value is $1,350. The average of these two is \(\frac{1350 + 1350}{2}=\frac{2700}{2}=1350\).

Answer:

\(1350\)