Answer to Question #225870 in Statistics and Probability for Chamali Kaluarachc

Question

A sample of 50 observations is taken from a normal population, with mu=100 and =10. If the population is finite with N=250. Find: a) 𝑃(𝑋̅ > 103) b) 𝑃(98 < 𝑋̅ < 101) c) Repeat above exercise with N=500

Answer 1

Use the Finite Correction Factor for $n>0.05N$

a) $\mu=100, \sigma=10, N=250, n=50$

$\dfrac{\sigma}{\sqrt{n}}\sqrt{\dfrac{N-n}{N-1}}=\dfrac{10}{\sqrt{50}}\sqrt{\dfrac{250-50}{250-1}}=\dfrac{20}{\sqrt{249}}$

$P(X>103)=1-P(Z\leq\dfrac{103-100}{\dfrac{20}{\sqrt{249}}})$

$\approx1-P(Z\leq2.367)\approx0.0090$

b) $\mu=100, \sigma=10, N=250, n=50$

$\dfrac{\sigma}{\sqrt{n}}\sqrt{\dfrac{N-n}{N-1}}=\dfrac{10}{\sqrt{50}}\sqrt{\dfrac{250-50}{250-1}}=\dfrac{20}{\sqrt{249}}$

$P(98<X<101)=P(X<101)-P(X\leq98)$

$=P(Z<\dfrac{101-100}{\dfrac{20}{\sqrt{249}}})-P(Z\leq\dfrac{98-100}{\dfrac{20}{\sqrt{249}}})$

$\approx P(Z<0.7890)-P(Z\leq-1.5780)$

$\approx0.784944-0.057283\approx0.7277$

c) $\mu=100, \sigma=10, N=500, n=50$

$\dfrac{\sigma}{\sqrt{n}}\sqrt{\dfrac{N-n}{N-1}}=\dfrac{10}{\sqrt{50}}\sqrt{\dfrac{500-50}{500-1}}=\dfrac{30}{\sqrt{499}}$

$P(X>103)=1-P(Z\leq\dfrac{103-100}{\dfrac{30}{\sqrt{499}}})$

$\approx1-P(Z\leq2.2338)\approx0.0127$

$P(98<X<101)=P(X<101)-P(X\leq98)$

$=P(Z<\dfrac{101-100}{\dfrac{30}{\sqrt{499}}})-P(Z\leq\dfrac{98-100}{\dfrac{30}{\sqrt{499}}})$

$\approx P(Z<0.7446)-P(Z\leq-1.4892)$

$\approx0.771746-0.068215\approx0.7035$