Question

You will receive a set of ten quantitative values for each of three genotypes: aa, Aa, and AA.

Answer 1

Mean for the Geno Code 0 is average of those ten values ((-3.97)+(-3.99) +(-5.55)+ (-4.56) + (-4.51) + (-4.2 )+( -4.95) + (-4.56) + (-4 )+ (-5.68))/10= -4.597

Standard deviation formula is √((-3.97- (-4.597)2 )+(-3.99-(-4.597)2) +(-5.55-(-4.597)2)+ (-4.56-(-4.597)2) + (-4.51-(-4.597)2) + (-4.2-(-4.597) 2 )+( -4.95-(-4.597)2) + (-4.56-(-4.597)2) + (-4 -(-4.597)2)+ (-5.68-(-4.597)2))/10= √  (0.39+0.37+0.91+0.001+0.007+0.16+0.12+0.001+0.36+1.17)/10= √0.3489=0.59

Mean for the Geno 1 is -0.25 (calculated on the same way as for Geno 0 with different numbers)  and standard deviation for Geno 1 is 0.65.

Mean for Geno 2 is 5.241 and standard deviation is 1.038.

2.

If individual Phenotype_Value is 5.11, the corresponding genotype code is Geno code 2 since the value is in the range of values 5.241+- 1.

If individual Phenotype_Value is 5.03 the corresponding genotype code is also Geno code 2.

Since the standard deviation is used to measure the variation (dispersal) of set of data values (how close the values are to the mean value) we can conclude whether are random value is within a certain range.