Question

consider the following sample data, which represent weights of sea mussels grown on the california coast, in grams: { 1.4, 1.8, 1.8, 1.9, 1.9, 1.9, 2, 2, 2.1, 2.2, 2.3, 2.3, 2.3, 2.3, 2.5, 2.5, 2.5, 2.6, 2.6 }. first, give the mean of the data set. part 2 of 5 next, give the median of the data set. part 3 of 5 now give the mode of the data set. if there is more than one, write them in order, separated by commas. part 4 of 5 give the midrange of the data set. part 5 of 5 given the relationship between the mean and median above, what shape is the distribution likely to be? the distribution will be roughly symmetric. the distribution will probably be skewed to the right. the distribution will probably be skewed to the left.

Explanation:

Step1: Calculate the mean

The formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are the data - points and $n$ is the number of data - points. Here, $n = 20$, and $\sum_{i=1}^{20}x_{i}=1.4 + 1.8\times2+1.9\times3 + 2\times2+2.1+2.2+2.3\times4+2.5\times3+2.6\times2=44$. So, $\bar{x}=\frac{44}{20}=2.2$.

Step2: Calculate the median

Since $n = 20$ (an even number), the median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th ordered data - points. The ordered data set has 20 values. The 10th value is 2.2 and the 11th value is 2.3. So, the median $M=\frac{2.2 + 2.3}{2}=2.25$.

Step3: Identify the mode

The mode is the most frequently occurring value. In the data set, 2.3 occurs 4 times, more frequently than any other value, so the mode is 2.3.

Step4: Calculate the mid - range

The mid - range is calculated as $\frac{\text{min}+\text{max}}{2}$. Here, $\text{min}=1.4$ and $\text{max}=2.6$. So, the mid - range $=\frac{1.4 + 2.6}{2}=2$.

Step5: Determine the shape of the distribution

The mean $\bar{x}=2.2$ and the median $M = 2.25$. Since the mean is less than the median, the distribution will probably be skewed to the left.

Answer:

Part 1: 2.2
Part 2: 2.25
Part 3: 2.3
Part 4: 2
Part 5: The distribution will probably be skewed to the left.