Question

consider the following sample data, which represent weights of sea mussels grown on the california coast, in grams: { 1.4, 1.5, 1.7, 2, 2.1, 2.1, 2.1, 2.1, 2.2, 2.2, 2.3, 2.3, 2.3, 2.4, 2.4, 2.5, 2.5, 2.7, 2.9 }. 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 probably be skewed to the left. the distribution will probably be skewed to the right. the distribution will be roughly symmetric.

Explanation:

Step1: Recall mean formula

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are data - points and $n$ is the number of data - points. Here $n = 20$, and $\sum_{i=1}^{20}x_{i}=1.4 + 1.5+1.7 + 2+2.1\times4+2.2\times2+2.3\times3+2.4\times2+2.5\times2+2.7+2.9=41.7$. So $\bar{x}=\frac{41.7}{20}=2.085$.

Step2: Recall median formula

For $n = 20$ (an even number of data - points), 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 $n = 20$ values. The 10th and 11th values are both $2.2$, so the median $M=\frac{2.2 + 2.2}{2}=2.2$.

Step3: Recall mode formula

The mode is the most frequently occurring value. The value $2.1$ appears 4 times, more frequently than any other value, so the mode is $2.1$.

Step4: Recall mid - range formula

The mid - range is calculated as $\frac{\text{Minimum value}+\text{Maximum value}}{2}$. The minimum value is $1.4$ and the maximum value is $2.9$. So the mid - range $=\frac{1.4 + 2.9}{2}=2.15$.

Step5: Analyze distribution shape

Since the mean ($2.085$) is less than the median ($2.2$), in a left - skewed distribution, the tail on the left pulls the mean in that direction. So the distribution is likely skewed to the left.

Answer:

Part 1: $2.085$
Part 2: $2.2$
Part 3: $2.1$
Part 4: $2.15$
Part 5: The distribution will probably be skewed to the left.