the weights (to the nearest five pounds) of 32 randomly selected male college students are organized in the…

Explanation:

Step1: Determine position of median

For $n = 32$ (even - numbered data set), the median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+ 1)$th ordered data - values. $\frac{n}{2}=\frac{32}{2}=16$ and $\frac{n}{2}+1 = 17$. So, we need to find the 16th and 17th ordered data - values.

Step2: Cumulative frequency calculation

Starting from the left - hand side of the histogram:

• For the weight interval 100 - 105, frequency $f_1=1$. Cumulative frequency $CF_1 = 1$.
• For 105 - 110, $f_2 = 2$, $CF_2=1 + 2=3$.
• For 110 - 115, $f_3 = 3$, $CF_3=3 + 3=6$.
• For 115 - 120, $f_4 = 4$, $CF_4=6 + 4 = 10$.
• For 120 - 125, $f_5 = 7$, $CF_5=10 + 7=17$. The 16th and 17th values fall within the 120 - 125 interval.

Step3: Assume uniform distribution within the interval

We assume that the data is uniformly distributed within the 120 - 125 interval. The lower limit of the interval $L = 120$, the frequency of the interval $f = 7$, the cumulative frequency of the previous interval $CF = 10$, and the class width $w=5$. The formula for the median of grouped data is $M = L+\frac{\frac{n}{2}-CF}{f}\times w$. Substituting the values: $\frac{n}{2}=16$, $CF = 10$, $f = 7$, $L = 120$, $w = 5$. $M=120+\frac{16 - 10}{7}\times5=120+\frac{30}{7}\approx120 + 4.29=124.29$. But since the weights are to the nearest five pounds, we consider the mid - point of the interval where the median lies. The mid - point of the 120 - 125 interval is 122.5.

Answer:

122.5