Please explain a What percentage 001 of flies have thorax le

Please explain

a/ What percentage (±0.01) of flies have thorax lengths less than 0.75 mm?____________ %

b/ What percentage (±0.01) of flies have thorax lengths greater than 0.85 mm?___________ %

c/ What percentage (±0.01) of flies have thorax lengths between 0.75 and 0.85 mm?__________ %

Solution

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.75      
u = mean =    0.81      
          
s = standard deviation =    0.084      
          
Thus,          
          
z = (x - u) / s =    -0.714285714      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.714285714   ) =    0.237525262 or 23.7525262% [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.85      
u = mean =    0.81      
          
s = standard deviation =    0.084      
          
Thus,          
          
z = (x - u) / s =    0.476190476      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.476190476   ) =    0.316969342 or 31.6969342% [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.75      
x2 = upper bound =    0.85      
u = mean =    0.81      
          
s = standard deviation =    0.084      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.714285714      
z2 = upper z score = (x2 - u) / s =    0.476190476      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.237525262      
P(z < z2) =    0.683030658      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.445505396 or 44.5505396 [ANSWER]