那么平均值Y~N（μ,σ^2/14）
1)σ^2=25 Y~N（μ,25/14）
那么P(|Y-μ|<=1.5)
=P(μ-1.5
=Φ((μ+1.5-μ)/√25/14)-Φ((μ-1.5-μ)/√25/14)
=Φ(3√14/10)-Φ(-3√14/10)
=2Φ(3√14/10)-1
=2Φ(1.12)-1查表Φ(1.12)=0.8686
=0.7372
(2)σ^2未知,但s^2=17.26
T=√n (Y-μ)/S ~t(n-1)分布
那么
P(|Y-μ|<=1.5)
=P(-1.5<=Y-μ<=1.5)
=P(-1.5*√n/S<=√n (Y-μ)/S<=1.5*√n/S)
=P(-1.35<=T<=1.35)查t分布表,發(fā)現(xiàn)P(|t(13)|>1.35)=0.2
=1-0.2
=0.8