Question 331603
Given Normally distributed data with mean=27.0 and standard deviation =1.3
1 standard deviation = 1.3
2 standard deviations=2.6
3 standard deviations =3.9
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A) The percentage of women with forward grip reaches between 24.4 inches and 29.6 inches.

compute the mid-range
Upper value-Xbar =29.6-27.0=2.6  
Xbar - Lower value =27.0-24.4=2.6
This means that 24.4 to 29.6 is 2 standard deviations from the mean
approximately 95% of the data will lie within 2 standard dev from the mean
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B) Find the percentage of women with forward grip reaches less than 30.9 inches.
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compute the distance in standard deviations from the mean to the largest value

30.9-27.0 = 3.9  this is 3 standard deviations above the mean
1/2 of 99.7% of the data will lie within the mean and 3 standard deviations 
50% of the data will lie below the mean
so, the answer is  50% + 1/2*99.7% =99.85%
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C) Find the percentage of women with forward grip reaches between 27.0 inches and 28.3 inches.
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compute the distance which starts at the mean 27.0 and goes to 28.3
28.3-27.0=1.3  this is one standard deviation above the mean

Since approx 68% of the data lies within 1 standard deviation on either side of the maan, the answer is 1/2* 68% = 34%