the weights (to the nearest five pounds) of 38 randomly selected male college students are organized in the…

Explanation:

Step1: Determine mid - points of each class

For the class 100 - 105, mid - point $x_1=\frac{100 + 105}{2}=102.5$. For 105 - 110, $x_2=\frac{105+110}{2}=107.5$, for 110 - 115, $x_3=\frac{110 + 115}{2}=112.5$, for 115 - 120, $x_4=\frac{115+120}{2}=117.5$, for 120 - 125, $x_5=\frac{120 + 125}{2}=122.5$, for 125 - 130, $x_6=\frac{125+130}{2}=127.5$, for 130 - 135, $x_7=\frac{130 + 135}{2}=132.5$, for 135 - 140, $x_8=\frac{135+140}{2}=137.5$, for 140 - 145, $x_9=\frac{140 + 145}{2}=142.5$, for 145 - 150, $x_{10}=\frac{145+150}{2}=147.5$, for 150 - 155, $x_{11}=\frac{150 + 155}{2}=152.5$.

Step2: Determine frequencies of each class

From the histogram, frequencies $f_1 = 1,f_2=3,f_3 = 4,f_4=6,f_5 = 8,f_6=9,f_7=4,f_8=2,f_9=1,f_{10}=0,f_{11}=1$.

Step3: Calculate the product $f_ix_i$ for each class

$f_1x_1=1\times102.5 = 102.5$, $f_2x_2=3\times107.5 = 322.5$, $f_3x_3=4\times112.5 = 450$, $f_4x_4=6\times117.5 = 705$, $f_5x_5=8\times122.5 = 980$, $f_6x_6=9\times127.5 = 1147.5$, $f_7x_7=4\times132.5 = 530$, $f_8x_8=2\times137.5 = 275$, $f_9x_9=1\times142.5 = 142.5$, $f_{10}x_{10}=0\times147.5 = 0$, $f_{11}x_{11}=1\times152.5 = 152.5$.

Step4: Calculate the sum of $f_ix_i$ and the sum of frequencies

$\sum_{i = 1}^{11}f_ix_i=102.5+322.5 + 450+705+980+1147.5+530+275+142.5+0+152.5=4800$. $\sum_{i=1}^{11}f_i=1 + 3+4+6+8+9+4+2+1+0+1=38$.

Step5: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}f_ix_i}{\sum_{i=1}^{n}f_i}=\frac{4800}{38}\approx126.316$.

Answer:

126.316