Question

given the following high temperatures from each month, fill in the blanks below! month temp january 63° february 64° march 68° april 74° may 80° june 84° july 86° august 86° september 84° october 78° november 70° december 48° the mean is (round to 2 decimals) the median is the mode is and the range is the minimum is q1 is q3 is the maximum is iqr is there are outliers

Explanation:

Step1: Calculate the mean

Sum of all temperatures: $63 + 64+68 + 74+80+84+86+86+84+78+70+48=865$
Number of data - points $n = 12$
Mean $\bar{x}=\frac{865}{12}\approx72.08$

Step2: Find the median

Arrange data in ascending order: $48,63,64,68,70,74,78,80,84,84,86,86$
Since $n = 12$ (even), median is the average of the $\frac{n}{2}$ - th and $(\frac{n}{2}+1)$ - th values.
Median $=\frac{74 + 78}{2}=76$

Step3: Determine the mode

The mode is the value that appears most frequently. Here, $84$ and $86$ both appear twice, so the modes are $84$ and $86$

Step4: Calculate the range

Range = Maximum - Minimum
Maximum = $86$, Minimum = $48$
Range = $86 - 48=38$

Step5: Find Q1 and Q3

The lower half of the data is $48,63,64,68,70,74$
$Q1$ is the median of the lower - half. Since there are 6 values (even), $Q1=\frac{64 + 68}{2}=66$
The upper half of the data is $78,80,84,84,86,86$
$Q3$ is the median of the upper - half. Since there are 6 values (even), $Q3=\frac{84+84}{2}=84$

Step6: Calculate the IQR

IQR = $Q3 - Q1$
IQR = $84 - 66 = 18$

Step7: Check for outliers

Lower fence = $Q1-1.5\times IQR=66-1.5\times18=66 - 27 = 39$
Upper fence = $Q3 + 1.5\times IQR=84+1.5\times18=84 + 27 = 111$
Since all data points are within the range $39$ to $111$, there are $0$ outliers

Answer:

The Mean is $72.08$
The Median is $76$
The Mode is $84$ and $86$
The Range is $38$
The Minimum is $48$
Q1 is $66$
Q3 is $84$
The Maximum is $86$
IQR is $18$
There are $0$ outliers