In a binomial distribution 31% of the items are under 45 and 8% are over 64. Find the mean and variance of the distribution
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The problem stated that XX is a random variable normally distributed with unknown mean and variance, but known quantiles q1=0.31q1=0.31 and q2=1–0.08q2=1–0.08. Therefore, we can write
P(X<45)=0.31P(X<45)=0.31
P(X>64)=0.08⟹P(X<64)=1–0.08=0.92P(X>64)=0.08⟹P(X<64)=1–0.08=0.92
It can be shown that P(X<x)=P(X−μσ<x−μσ)=P(Z<z)P(X<x)=P(X−μσ<x−μσ)=P(Z<z), where Z∼N(0,1)Z∼N(0,1) - i.e. the standard Normal distribution - and xx is a realisation of the random variable X.X. Therefore, we can write
P(X−μσ<45−μσ)=P(Z<z1)=Φ(z1)=0.31P(X−μσ<45−μσ)=P(Z<z1)=Φ(z1)=0.31
P(X−μσ<64−μσ)=P(Z<z2)=Φ(z2)=0.92,P(X−μσ<64−μσ)=P(Z<z2)=Φ(z2)=0.92,
where z1=45−μσz1=45−μσ and z_2=64−μσ64−μσ.
From the quantile function of the standard Normal distribution we can find z1z1and z2,z2,e.g. in R use qnorm(0.31) and qnorm(0.92)to get the desired quantiles z1z1 and z2z2. It turns out z1=−0.4958503z1=−0.4958503 and z2=1.405072z2=1.405072. We can now write the following system
z1=45−μσ=−0.4958503z1=45−μσ=−0.4958503
z2=64−μσ=1.405072z2=64−μσ=1.405072
This is a system with two equations and two variables, therefore we have a unique solution, μ=49.9561μ=49.9561 and σ=9.995148σ=9.995148.
You may obtain slightly different results since Φ(z)Φ(z) is an approximation and differ from one software to another.
P(X<45)=0.31P(X<45)=0.31
P(X>64)=0.08⟹P(X<64)=1–0.08=0.92P(X>64)=0.08⟹P(X<64)=1–0.08=0.92
It can be shown that P(X<x)=P(X−μσ<x−μσ)=P(Z<z)P(X<x)=P(X−μσ<x−μσ)=P(Z<z), where Z∼N(0,1)Z∼N(0,1) - i.e. the standard Normal distribution - and xx is a realisation of the random variable X.X. Therefore, we can write
P(X−μσ<45−μσ)=P(Z<z1)=Φ(z1)=0.31P(X−μσ<45−μσ)=P(Z<z1)=Φ(z1)=0.31
P(X−μσ<64−μσ)=P(Z<z2)=Φ(z2)=0.92,P(X−μσ<64−μσ)=P(Z<z2)=Φ(z2)=0.92,
where z1=45−μσz1=45−μσ and z_2=64−μσ64−μσ.
From the quantile function of the standard Normal distribution we can find z1z1and z2,z2,e.g. in R use qnorm(0.31) and qnorm(0.92)to get the desired quantiles z1z1 and z2z2. It turns out z1=−0.4958503z1=−0.4958503 and z2=1.405072z2=1.405072. We can now write the following system
z1=45−μσ=−0.4958503z1=45−μσ=−0.4958503
z2=64−μσ=1.405072z2=64−μσ=1.405072
This is a system with two equations and two variables, therefore we have a unique solution, μ=49.9561μ=49.9561 and σ=9.995148σ=9.995148.
You may obtain slightly different results since Φ(z)Φ(z) is an approximation and differ from one software to another.
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