Question

1. 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?

a. the mean must be smaller than the median because the data set includes more negative values than positive values.
b. because q1 is equal to 6 and q3 is equal to 8, the iqr is equal to 2.
c. changing the highest temperature from a 12 to a -12 will lower both the mean and the median.
d. the standard deviation of the data set will be negative because more than half of the data values are negative.
e. 25% of the temperatures in this data set are below -5.

Explanation:

Step1: Arrange data in ascending order

$-10,-5,-3,-2,-1,0,4,6,12$

Step2: Calculate the mean

The mean $\bar{x}=\frac{-10 + 6+0 - 5-3 - 1+4+12 - 2}{9}=\frac{-9}{9}=-1$

Step3: Find the median

Since $n = 9$ (odd), the median is the 5 - th value. So the median is $-1$.

Step4: Analyze option C

Original data: $-10,-5,-3,-2,-1,0,4,6,12$. Mean $=-1$, median $=-1$.
New data (after changing 12 to - 12): $-12,-10,-5,-3,-2,-1,0,4,6$.
New mean $\bar{x}=\frac{-12-10 - 5-3 - 2-1+0+4+6}{9}=\frac{-23}{9}\approx - 2.56$.
New median: Since $n = 9$ (odd), the median is the 5 - th value, which is $-2$. Both the mean and median are lowered.

Step5: Analyze other options

• Option A: Just because there are more negative values doesn't mean mean < median. Here mean = median originally.
• Option B: First, find quartiles. After arranging data: $-10,-5,-3,-2,-1,0,4,6,12$. $Q_1$ is the median of lower half $(-10,-5,-3,-2,-1)$, so $Q_1=-3$. $Q_3$ is the median of upper half $(0,4,6,12)$, so $Q_3 = 5$. $IQR=Q_3 - Q_1=5-(-3)=8$.
• Option D: Standard deviation $\sigma=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n}}$, and it is non - negative.
• Option E: After arranging data, 2 out of 9 values are below - 5. $\frac{2}{9}\approx22.2\%

eq25\%$.

Answer:

C. Changing the highest temperature from a 12 to a -12 will lower both the mean and the median.