Question

question 1 1 pts in minneapolis, the low temperatures in degrees fahrenheit for nine days in february are as follows: -10, 6, 0, -5, -3, -1, 4, 12, and -2. which one of the following statements about this data set is correct? the mean must be smaller than the median because the data set includes more negative values than positive values. because q1 is equal to 6 and q3 is equal to 8, the iqr is equal to 2. changing the highest temperature from a 12 to a -12 will lower both the mean and the median. the standard deviation of the data set will be negative because more than half of the data values are negative. 25% of the temperatures in this data set are below -5.

Explanation:

Step1: Sort the data set

The sorted data set is: $-10, -5, -3, -2, -1, 0, 4, 6, 12$.

Step2: Analyze each statement

Mean - Median comparison

The mean $\bar{x}=\frac{- 10 + 6+0 - 5-3 - 1+4 + 12-2}{9}=\frac{-9}{9}=-1$. The median of 9 - data set (odd - numbered) is the 5th value, which is $-1$. So the mean is not smaller than the median just because there are more negative values.

Quartile and IQR

For a 9 - data set, the position of $Q_1$ is $\frac{n + 1}{4}=\frac{9+1}{4}=2.5$. So $Q_1=\frac{-5-3}{2}=-4$. The position of $Q_3$ is $\frac{3(n + 1)}{4}=\frac{3\times(9 + 1)}{4}=7.5$. So $Q_3=\frac{4 + 6}{2}=5$. $IQR=Q_3 - Q_1=5-(-4)=9$.

Changing the highest value

Original mean $\bar{x}=-1$. New mean with 12 changed to - 12: $\bar{x}_{new}=\frac{-10 + 6+0 - 5-3 - 1+4 - 12-2}{9}=\frac{-23}{9}\approx - 2.56$. The original median is $-1$. After changing, the sorted data is $-12,-10, -5, -3, -2, -1, 0, 4, 6$. The new median is still $-1$.

Standard deviation

Standard deviation $\sigma=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n}}$. Since it is a square - root of a non - negative quantity, it is non - negative.

Percentile

Since $Q_1=-4$, 25% of the data is below $Q_1$. And $-5$ is below $Q_1$, so 25% of the temperatures in this data set are below $-5$.

Answer:

25% of the temperatures in this data set are below -5.