Normal Distributions Essay

Question#1

Tommy is 3.5 months old with weight 6.41Kg. The normal distribution corresponding to his age group has mean of 6.39Kg and standard deviation of 0.12. The number of male persons who weigh more than him is calculated by using the relation:
Px<z=0.5(1+erfz2)
=0.5(erfc-z2)
Where “z” is z-score, “x” is the normal distribution variable, “erf” is error function and “erfc” is the complementary error function. The latter equation will help us in finding the area under Gaussian curve for particular z_score in MATLAB.
z=6.41-6.390.12=0.16666666
Px>z=1-Px<z=1-0.5*erfc*(-0.1666662)
=1-0.5662=43.35%

Question 2

Paul is 11.5 months old and his weight is 10.02Kg. In order to find the percentage of males who weigh more than him, we follow the same procedure as mentioned in Question 1.
z=10.02-10.160.11=-1.27272727
Px>z=1-Px<z=1-0.5*erfc*(1.272727272)
=1-0.1016=89.54%

Question 3

As obvious, the mentioned patient scores more than the mean value. Therefore, the percentage of patients who score more than him will be less than 50%. The formulas for finding the exact value are the same as discussed previously.
z=54-525=0.4
Px>z=1-Px<z=1-0.5*erfc*(-0.42)
=1-0.6554=34.46%

Question 4

The child’s percentile rank is the number of students who scored less than him. This can be calculated very easily by finding the area under Gaussian curve that lies to the left of child’s score.
z=148-12515=1.53333
Percentile= Px<z=0.5*erfc*(-1.5333332)
=93.74%

Question 5

zPAT=76-808=-0.5
zCHRIS=94-808=1.75
Px>zpat&x<zCHRIS=Px<zCHRIS-P(x<zPAT)
=0.5*erfc*-1.752-0.5*erfc*0.52
=0.9599-0.3055
=65.44%
Therefore, 65.44% students will score between Pat and Chris.

References

Stanley, W. D. (2004). Technical Analysis and Applications with MATLAB. Cengage Learning.