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160 athletes are distributed with a mean finishing time of 72.2 min and standard deviation of 6.8 min

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In a standard bell curve, the mean finishing time is the middle of the curve. One standard deviation includes 68.2% of the values with 34.1% above and 34.1% below the mean. For two deviations it includes 95.4% of the values with 47.7% above and 47.7% below the mean.
 
In this problem 72.2 is the mean and 6.8 the standard deviation. One standard deviation below would include the range of values between 72.2 and 65.4(=72.2-6.8) the number of people in this range would be 34.1% of the total number of people. In this case its two standard deviations below which includes the range between 72.2 and 58.6(=72.2-2*6.8). This includes 47.7% of the total number of athletes who participated. So the answer would be 160*.477=76.32 or 76 since only counting whole people.
Hi Drew,
 
Mean=72.2min, standard deviation=6.8
# of Athletes with finishing times between 58.6 & 72.2.
 
x∼N(72.2,6.8)
P(58.6<x<72.2)= P(58.6-72.2/6.8 <x-μ/σ <72.2-72.2/6.8)
                        = P(-2<Z<0)
                        = P(0)-P( -2)
                        =0.5000-0.0228......              (from Standard normal probability table)
                        =0.4772
so 47.72% students bet 58.6 min to 72.2 min
 
Total # of Athletes=160
# of atheletes with finishing time bet 58.6 & 72.2 = 160*47.72/100= 76.352
                                                                       =76. 
Meena from Strongsville, OH