Question

a used - car salesman has 28 cars in his inventory, with prices ranging from $11,500 to $25,000. for a labor day sale, he reduces the price of each car by $500. which of the following choices correctly describes the effect this sale will have on each of the following characteristics of the resulting distribution of price: shape, mean and median, and standard deviation and interquartile range (iqr)?
shape: unchanged
mean and median: unchanged
standard deviation and interquartile range (iqr): unchanged
shape: become more skewed to the left
mean and median: decrease by $500
standard deviation and interquartile range (iqr): unchanged
shape: unchanged
mean and median: decrease by $500
standard deviation and interquartile range (iqr): unchanged
shape - unchanged
mean and median: decrease by $500
standard deviation and interquartile range (iqr): decrease by $500

Explanation:

Step1: Effect on shape

When a constant ($500$) is subtracted from each data - point, the relative positions of the data - points with respect to each other remain the same. So, the shape of the distribution (which is determined by the relative spread and position of data - points) is unchanged.

Step2: Effect on mean and median

The formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. If we define $y_{i}=x_{i}-500$, then the new mean $\bar{y}=\frac{\sum_{i = 1}^{n}(x_{i}-500)}{n}=\frac{\sum_{i = 1}^{n}x_{i}-500n}{n}=\bar{x}-500$.

For the median, which is the middle value (when $n$ is odd) or the average of the two middle values (when $n$ is even) of the ordered data set. If we subtract $500$ from each data - point, the middle value (or the average of the two middle values) of the new ordered data set $y_{i}=x_{i}-500$ will be the median of the original data set minus $500$.

Step3: Effect on standard deviation and IQR

The formula for the standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$. For $y_{i}=x_{i}-500$, the new mean $\bar{y}=\bar{x}-500$. Then $(y_{i}-\bar{y})=(x_{i}-500)-(\bar{x}-500)=x_{i}-\bar{x}$. So, $s_{y}=\sqrt{\frac{\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}{n - 1}}=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}=s_{x}$.

The inter - quartile range $IQR = Q_{3}-Q_{1}$. If $y_{i}=x_{i}-500$, then $Q_{3y}=Q_{3x}-500$ and $Q_{1y}=Q_{1x}-500$. So, $IQR_{y}=(Q_{3x}-500)-(Q_{1x}-500)=Q_{3x}-Q_{1x}=IQR_{x}$

Answer:

Shape: unchanged; Mean and median: decrease by $500$; Standard deviation and interquartile range ($IQR$): unchanged. So the correct option is the third one.