\begin{tabular}{|c|c|}
\hline Sample & Sample mean \\
\hline 1 & 16.8 \\
\hline 2 & 12.3 \\
\hline 3 & 19.0 \\
\hline 4 & 17.5 \\
\hline 5 & 18.2 \\
\hline 6 & 17.5 \\
\hline
\end{tabular}

Looking at the means of the samples, which of these values is least likely to be the mean of the population the samples were taken from?

A. 12.0
B. 18.4
C. 19.0
D. 19.5



Answer :

To determine which of the given values is least likely to be the mean of the population from which the samples were taken, we need to analyze the deviations. Here’s how we can approach this step-by-step:

1. Calculate the Sample Mean:
We are given the sample means as:
[tex]\[ 16.8, \, 12.3, \, 19.0, \, 17.5, \, 18.2, \, 17.5 \][/tex]
To find the average (mean) of these sample means, we sum them up and divide by the number of samples:
[tex]\[ \text{Sample mean} = \frac{16.8 + 12.3 + 19.0 + 17.5 + 18.2 + 17.5}{6} \][/tex]
[tex]\[ \text{Sample mean} = \frac{101.3}{6} \approx 16.8833 \][/tex]

2. Calculate Deviations:
Next, we compare this sample mean to the given population mean candidates (12, 18.4, 19.0, and 19.5) by calculating the absolute deviation for each:
[tex]\[ \begin{align*} \text{Deviation from 12} & = |16.8833 - 12| \approx 4.8833 \\ \text{Deviation from 18.4} & = |16.8833 - 18.4| \approx 1.5167 \\ \text{Deviation from 19.0} & = |16.8833 - 19.0| \approx 2.1167 \\ \text{Deviation from 19.5} & = |16.8833 - 19.5| \approx 2.6167 \end{align*} \][/tex]

3. Identify the Maximum Deviation:
The absolute deviations are:
[tex]\[ 4.8833, \, 1.5167, \, 2.1167, \, 2.6167 \][/tex]
The maximum deviation among these is 4.8833.

4. Find the Least Likely Population Mean:
The population mean corresponding to the maximum deviation is the least likely to be the actual population mean. Thus, the least likely mean of the population is 12.

In conclusion:
[tex]\[ \boxed{12} \][/tex]