Question

hint 1 hint 2 hint 3 hint 4 hint 5
below are iq scores from 30 randomly selected adults.
{ 83, 85, 88, 89, 91, 92, 93, 95, 96, 96, 97, 100, 100, 101, 102, 103, 104, 106, 106, 109, 110, 110, 111, 111, 112, 120, 122, 122, 122, 130 }.
first, give the mean of the data set.
104.2
part 2 of 6
next, give the median of the data set.
102
part 3 of 6
now give the mode of the data set. if there is more than one, write them in order, separated by commas.
122
part 4 of 6
finally, give the midrange of the data set.
106.5
part 5 of 6
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 right.
the distribution will probably be skewed to the left.
the distribution will be roughly symmetric.

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $n = 30$ and $x_{i}$ are the data - points. $\sum_{i=1}^{30}x_{i}=83 + 85+88+\cdots+130=3126$, so $\bar{x}=\frac{3126}{30}=104.2$.

Step2: Calculate the median

Since $n = 30$ (an even number), the median is the average of the $\frac{n}{2}$ - th and $(\frac{n}{2}+1)$ - th ordered data - points. The 15th and 16th ordered data - points are 102 and 103, so the median $=\frac{102 + 103}{2}=102.5$.

Step3: Determine the mode

The mode is the data - point that appears most frequently. In the data set, 122 appears 3 times, more frequently than any other number, so the mode is 122.

Step4: Calculate the mid - range

The mid - range is calculated as $\frac{\text{Minimum value}+\text{Maximum value}}{2}$. The minimum value is 83 and the maximum value is 130, so the mid - range $=\frac{83 + 130}{2}=106.5$.

Step5: Analyze the shape of the distribution

The mean ($\bar{x}=104.2$) is greater than the median ($102.5$). In a distribution, when the mean is greater than the median, the distribution is likely to be skewed to the right.

Answer:

Part 1: 104.2
Part 2: 102.5
Part 3: 122
Part 4: 106.5
Part 5: The distribution will probably be skewed to the right.