Question

If the height of the soldiers in the army of a country follows normal distribution with mean of 70 inches and the standard deviation of 5.3 inches. If 900 soldersare selected at random from this army then find the number of solders havingtheirheights will
a) At least 62 inches
b) B/W 68 to 75 inches
c) Below 74 inches
d) What value of the height will have35% of the solders above it

Answer 1

$\xi\text{ --- random variable (random value of height)}.\\ P\{\xi\geq 62\}=1-P\{\xi<62\}=1-F(62)=\\=1-\Phi(\frac{62-70}{5.3})= 1-\Phi(-1.509)=1-0.0656=0.9344.\\ F(x)=\frac{1}{{(5.3)}\sqrt{2\pi}}\int_{-\infty}^x e^{-\frac{(t-70)^2}{2\cdot({5.3})^2}}dt.\\ \Phi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x} e^{-\frac{t^2}{2}}dt.\\ \mu_{\overline{p}}=P\text{ (mean of sample proportions equals population proportion)}.\\ \text{So we have a) }(0.9344)\cdot 900\approx 840.\\ b) P\{68\leq\xi\leq 75\}=F(75)-F(68)=\Phi(\frac{75-70}{5.3})-\Phi(\frac{68-70}{5.3})\approx \\ \approx \Phi(0.943)-\Phi(-0.377)\approx 0.8272-0.3531=0.4741.\\ \text{So we have } P=(0.4741)\cdot 900\approx 426.\\ c) P\{\xi<74\}=F(74)=\Phi(\frac{74-70}{5.3})=\Phi(0.755)=0.7749.\\ \text{So we have } P=(0.7749)\cdot 900\approx 697.\\ d) P\{\xi>t\}=0.35.\\ \text{We should find }t.\\ P\{\xi>t\}=1-P\{\xi\leq t\}=0.35.\\ P\{\xi\leq t\}=0.65.\\ \Phi(\frac{t-70}{5.3})=0.65.\\ \frac{t-70}{5.3}=0.39.\\ t\approx 72.$