Questions: For the data in the table, find the mean and the median of the monthly high temperatures. Then do the same for the monthly low temperatures. Month High Low --- --- --- Jan 18 9 Feb 24 14 Mar 30 19 Apr 44 31 May 55 44 Temperatures are in degrees Fahrenheit The mean of the high temperatures is °F. (Type an integer or a decimal.) The median of the high temperatures is °F. (Type an integer or a decimal.) The mean of the low temperatures is °F. (Type an integer or a decimal.) The median of the low temperatures is °F. (Type an integer or a decimal.)

Solution Steps

To find the mean of the high temperatures, we use the formula:

\[ \mu = \frac{\sum_{i=1}^N x_i}{N} \]

where \( N \) is the number of data points and \( x_i \) are the individual temperature values. For the high temperatures:

\[ \mu = \frac{18 + 24 + 30 + 44 + 55}{5} = \frac{171}{5} = 34.2 \]

Thus, the mean of the high temperatures is \( 34.2 \, ^\circ \text{F} \).

To find the median, we first sort the high temperatures:

\[ \text{Sorted data: } [18, 24, 30, 44, 55] \]

Since there are 5 data points (an odd number), the median is the value at position \( \frac{N + 1}{2} = \frac{5 + 1}{2} = 3 \). Therefore, the median is:

\[ \text{Median} = 30 \, ^\circ \text{F} \]

Using the same mean formula for the low temperatures:

\[ \mu = \frac{\sum_{i=1}^N x_i}{N} \]

For the low temperatures:

\[ \mu = \frac{9 + 14 + 19 + 31 + 44}{5} = \frac{117}{5} = 23.4 \]

Thus, the mean of the low temperatures is \( 23.4 \, ^\circ \text{F} \).

We sort the low temperatures:

\[ \text{Sorted data: } [9, 14, 19, 31, 44] \]

Again, with 5 data points, the median is at position \( \frac{N + 1}{2} = 3 \). Therefore, the median is:

\[ \text{Median} = 19 \, ^\circ \text{F} \]

Final Answer