the distance a student in pe class can run in a 20 minute time period is shown in the histogram below.\npe…

Explanation:

Step1: Count total number of students

Let's assume the frequencies for the intervals are (f_1,f_2,f_3,f_4,f_5,f_6,f_7) from left - to - right. Count the sum (N=\sum_{i = 1}^{7}f_i). If (N) is odd, the median is the (\left(\frac{N + 1}{2}\right))-th value, if (N) is even, the median is the average of the (\frac{N}{2})-th and (\left(\frac{N}{2}+1\right))-th values. By looking at the histogram, we can see that the cumulative frequency builds up and the middle value (median) falls in the (2.1 - 2.5) range, so statement #1 is False.

Step2: Estimate the mean

We estimate the mean (\bar{x}=\frac{\sum_{i = 1}^{7}x_if_i}{\sum_{i = 1}^{7}f_i}), where (x_i) is the mid - point of each interval. The mid - points for the intervals (0.1 - 0.5,0.6 - 1.0,1.1 - 1.5,1.6 - 2.0,2.1 - 2.5,2.6 - 3.0,3.1 - 3.4) are (0.3,0.8,1.3,1.8,2.3,2.8,3.25) respectively. Since the data has a right - skewed distribution (more data on the left side of the peak), the mean is pulled towards the right of the median. The mean is likely in the (2.1 - 2.5) range, so statement #2 is True.

Step3: Check for symmetry

A symmetric distribution has the same shape on both sides of the center. Looking at the histogram, the left - hand side of the peak (around (2.1 - 2.5)) has fewer data points compared to the right - hand side. So the distribution is not symmetric, and statement #3 is False.

Answer:

#1: False #2: True #3: False