Since heights are normally distributed, the probability P ( x ≤ x 0 ) P(x\leq x_0) P ( x ≤ x 0 ) is given by (see https://en.wikipedia.org/wiki/Normal_distribution#Standard_normal_distribution ):
P ( x ≤ x 0 ) = 1 σ 2 π ∫ − ∞ x 0 e − 1 2 ( x − μ σ ) 2 d x P(x\leq x_0)=\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{x_0}e^{-\frac12(\frac{x-\mu}{\sigma})^2}dx P ( x ≤ x 0 ) = σ 2 π 1 ∫ − ∞ x 0 e − 2 1 ( σ x − μ ) 2 d x
with μ = 144.5 \mu=144.5 μ = 144.5 , σ = 6 \sigma=6 σ = 6 . In a similar way, we can express the probabilities P ( x 1 ≤ x ≤ x 2 ) P(x_1\leq x\leq x_2) P ( x 1 ≤ x ≤ x 2 )
and P ( x ≥ x 3 ) P(x\geq x_3) P ( x ≥ x 3 ) for real x 0 , x 1 , x 2 , x 3 x_0,x_1,x_2,x_3 x 0 , x 1 , x 2 , x 3 by changing the boundaries in the integral. We obtained the following answers:
(i) P ( x ≤ 130 ) = 1 6 2 π ∫ − ∞ 130 e − 1 2 ( x − 144.5 6 ) 2 d x ≈ 0.0078 P(x\leq 130)=\frac{1}{6\sqrt{2\pi}}\int_{-\infty}^{130}e^{-\frac12(\frac{x-144.5}{6})^2}dx\approx0.0078 P ( x ≤ 130 ) = 6 2 π 1 ∫ − ∞ 130 e − 2 1 ( 6 x − 144.5 ) 2 d x ≈ 0.0078
(ii) P ( 151.5 ≤ x ≤ 182 ) = 1 6 2 π ∫ 151.5 182 e − 1 2 ( x − 144.5 6 ) 2 d x ≈ 0.1217 P(151.5\leq x\leq 182)=\frac{1}{6\sqrt{2\pi}}\int_{151.5}^{182}e^{-\frac12(\frac{x-144.5}{6})^2}dx\approx0.1217 P ( 151.5 ≤ x ≤ 182 ) = 6 2 π 1 ∫ 151.5 182 e − 2 1 ( 6 x − 144.5 ) 2 d x ≈ 0.1217
(iii)P ( x ≥ 150 ) = 1 6 2 π ∫ 150 + ∞ e − 1 2 ( x − 144.5 6 ) 2 d x ≈ 0.1797 P(x\geq 150)=\frac{1}{6\sqrt{2\pi}}\int_{150}^{+\infty}e^{-\frac12(\frac{x-144.5}{6})^2}dx\approx0.1797 P ( x ≥ 150 ) = 6 2 π 1 ∫ 150 + ∞ e − 2 1 ( 6 x − 144.5 ) 2 d x ≈ 0.1797
(iv) The aim is to find F − 1 ( 0.6 ) F^{-1}(0.6) F − 1 ( 0.6 ) , where F ( y ) = P ( x > y ) = 1 6 2 π ∫ y + ∞ e − 1 2 ( x − 144.5 6 ) 2 d x F(y)=P(x> y)=\frac{1}{6\sqrt{2\pi}}\int_y^{+\infty}e^{-\frac12(\frac{x-144.5}{6})^2}dx F ( y ) = P ( x > y ) = 6 2 π 1 ∫ y + ∞ e − 2 1 ( 6 x − 144.5 ) 2 d x
For y ≈ 143.0 y\approx143.0 y ≈ 143.0 (it was rounded to 1 decimal place) we get that F ( 143.0 ) ≈ 0.5987 F(143.0)\approx0.5987 F ( 143.0 ) ≈ 0.5987
For the computations we used the following program written in Anaconda (https://www.anaconda.com/products/individual ):
from scipy import integrate
import numpy as np
import math
func = lambda x:(1/(6*math.sqrt(2)*math.sqrt(math.pi)))*math.exp(-1/2*((x-144.5)/6)*((x-144.5)/6))
Pr1 = integrate.quad(func, -np.Infinity, 130)
Pr2 = integrate.quad(func, 151.5, 182)
Pr3 = integrate.quad(func, 150, +np.Infinity)
print(Pr1)
print(Pr2)
print(Pr3)
For the last task we used the fact that F ( y ) F(y) F ( y ) decreases and substituted different values of y y y until we got the values near 0.6.
Answers: (i) 0.0078 (ii) 0.1217 (iii) 0.1797 (iv) 143.0; the answers (i)-(iii) are rounded to 4 decimal places; (iv) is rounded to 1 decimal place.
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