\begin{tabular}{lllllllll}
1.48 & 1.56 & 1.56 & 1.59 & 1.60 & 1.61 & 1.63 & 1.64 & 1.64 \\
1.67 & 1.68 & 1.68 & 1.68 & 1.68 & 1.69 & 1.70 & 1.71 & 1.72 \\
1.72 & 1.75 & 1.78 & 1.76 & 1.77 & 1.77 & 1.79 & 1.81 & 1.81 \\
1.84 & 1.84 & 1.88 & 1.94 & & & & &
\end{tabular}



Answer :

Let's analyze the given data:

[tex]\[ 1.48, 1.56, 1.56, 1.59, 1.60, 1.61, 1.63, 1.64, 1.64, 1.67, 1.68, 1.68, 1.68, 1.68, 1.69, 1.70, 1.71, 1.72, 1.72, 1.75, 1.78, 1.76, 1.70, 1.77, 1.79, 1.81, 1.81, 1.84, 1.84, 1.88, 1.94 \][/tex]

### Step-by-Step Solution:

1. Calculating the Mean:
The mean (average) is determined by summing all the data points and then dividing by the number of data points.

Given the data set is:
[tex]\[ 1.48, 1.56, 1.56, 1.59, 1.60, 1.61, 1.63, 1.64, 1.64, 1.67, 1.68, 1.68, 1.68, 1.68, 1.69, 1.70, 1.71, 1.72, 1.72, 1.75, 1.78, 1.76, 1.70, 1.77, 1.79, 1.81, 1.81, 1.84, 1.84, 1.88, 1.94 \][/tex]

Summing these values, we have [tex]\( \sum = 52.909 \)[/tex]

The number of data points [tex]\( n = 31 \)[/tex].

The mean [tex]\( \bar{x} = \frac{\sum}{n} = \frac{52.909}{31} \approx 1.7067741935483876 \)[/tex].

Therefore, the mean is approximately 1.707.

2. Calculating the Median:
The median is the middle value in a data set that has been ordered from smallest to largest. If there is an even number of observations, the median is the average of the two middle numbers.

Ordered data:
[tex]\[ 1.48, 1.56, 1.56, 1.59, 1.60, 1.61, 1.63, 1.64, 1.64, 1.67, 1.68, 1.68, 1.68, 1.68, 1.69, 1.70, 1.70, 1.71, 1.72, 1.72, 1.75, 1.76, 1.77, 1.78, 1.79, 1.81, 1.81, 1.84, 1.84, 1.88, 1.94 \][/tex]

Since there are 31 data points (an odd number), the median is the value at position [tex]\( \frac{n+1}{2} = \frac{31+1}{2} = 16 \)[/tex].

The median value is at the 16th position, which is 1.70.

3. Calculating the Standard Deviation:
The standard deviation measures the amount of variation or dispersion of a set of values.

Using the formula for standard deviation for a sample:

[tex]\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \][/tex]

Applying this formula for our data set gives a standard deviation of approximately 0.10014141925375378.

### Summary:
- Mean (Average): 1.707
- Median: 1.70
- Standard Deviation: 0.100

These values give us a good understanding of the central tendency and dispersion of the given data set.