You constructed a genomic DNA library before you send it to

You constructed a genomic DNA library before you send it to sequencing. The length of DNA fragments follows Normal distribution with mean 450bp and standard deviation 200bp.

a) What is the proportion of DNA fragments longer than 800bp?

b) Calculate Z score first then use the Z table to find out the result. (1 point)

c) Then you did a size selection on the DNA library, only retain the fragments ranging from 400bp-500bp. Assume you did an ideal experiment with no extra loss. What percentage of DNA fragments you would have left after size selection? (1 point)

Solution

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    800      
u = mean =    450      
          
s = standard deviation =    200      
          
Thus,          
          
z = (x - u) / s =    1.75      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.75   ) =    0.040059157 [answer]

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b)

As above, z = 1.75, and using a z table,

P(z>1.75) = 0.0401 [answer]

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c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    400      
x2 = upper bound =    500      
u = mean =    450      
          
s = standard deviation =    200      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.25      
z2 = upper z score = (x2 - u) / s =    0.25      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.401293674      
P(z < z2) =    0.598706326      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.197412651 = 19.74% [ANSWER]