확률과 통계확률분포수능 기출기본 문제 (3점 중반)

정규분포표 확률

문제

정규분포 N(m,σ2){\mathrm{N}} \left( m , \sigma ^{2} \right)을 따르는 확률변수 XX에 대하여

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P(m≤X≤2m)=0.4772{\mathrm{P}} \left( m \leq X \leq 2 m \right) = 0.4772, P(X≥2)=0.8413{\mathrm{P}} \left( X \geq 2 \right) = 0.8413 일 때, P(0≤X≤5){\mathrm{P}} \left( 0 \leq X \leq 5 \right)의 값을 다음 표준정규분포표를 이용하여 구한 것은? [3점] ①0.53280.5328②0.62470.6247③0.66870.6687④0.68260.6826⑤0.77450.7745

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해설

확률변수 XX가 정규분포 N(m,σ2){\mathrm{N}} \left( m , \sigma ^{2} \right)을 따르므로 ZZ가 표준정규분포를 따르는 확률변수일 때 P(m≤X≤2m)=P(0≤Z≤mσ)=0.4772\displaystyle {\mathrm{P}} \left( m \leq X \leq 2 m \right) = {\mathrm{P}} \left( 0 \leq Z \leq \frac{m}{\sigma} \right) = 0.4772 주어진 표준정규분포표를 이용하면 P(0≤Z≤2)=0.4772{\mathrm{P}} \left( 0 \leq Z \leq 2 \right) = 0.4772이므로 mσ=2\displaystyle \frac{m}{\sigma} = 2, m=2σm = 2 \sigma P(X≥2){\mathrm{P}} \left( X \geq 2 \right)=P(Z≥2−mσ)\displaystyle = {\mathrm{P}} \left( Z \geq \frac{2 - m}{\sigma} \right)=P(Z≥2−2σσ)=0.8413\displaystyle = {\mathrm{P}} \left( Z \geq \frac{2 - 2 \sigma}{\sigma} \right) = 0.8413 그러므로 2−2σσ<0\displaystyle \frac{2 - 2 \sigma}{\sigma} < 0 P(X≥2){\mathrm{P}} \left( X \geq 2 \right)=P(2−2σσ≤Z≤0)+0.5\displaystyle = {\mathrm{P}} \left( \frac{2 - 2 \sigma}{\sigma} \leq Z \leq 0 \right) + 0.5 =P(0≤Z≤2σ−2σ)+0.5\displaystyle = {\mathrm{P}} \left( 0 \leq Z \leq \frac{2 \sigma - 2}{\sigma} \right) + 0.5 P(0≤Z≤2σ−2σ)=0.3413\displaystyle {\mathrm{P}} \left( 0 \leq Z \leq \frac{2 \sigma - 2}{\sigma} \right) = 0.3413 주어진 표준정규분포표를 이용하면 P(0≤Z≤1)=0.3413{\mathrm{P}} \left( 0 \leq Z \leq 1 \right) = 0.3413이므로 2σ−2σ=1\displaystyle \frac{2 \sigma - 2}{\sigma} = 1에서 σ=2\sigma = 2, m=4m = 4 P(0≤X≤5){\mathrm{P}} \left( 0 \leq X \leq 5 \right) =P(−2≤Z≤0.5)= {\mathrm{P}} \left( - 2 \leq Z \leq 0.5 \right) =P(−2≤Z≤0)+P(0≤Z≤0.5)= {\mathrm{P}} \left( - 2 \leq Z \leq 0 \right) + {\mathrm{P}} \left( 0 \leq Z \leq 0.5 \right) =P(0≤Z≤2)+P(0≤Z≤0.5)= {\mathrm{P}} \left( 0 \leq Z \leq 2 \right) + {\mathrm{P}} \left( 0 \leq Z \leq 0.5 \right) =0.4772+0.1915= 0.4772 + 0.1915 =0.6687= 0.6687

태그

#정규분포#표준화

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