Answer to Question #262759 in Statistics and Probability for yyy

Question #262759

A factory produces apple juice contained in a bottle of 1.5L. However, due to random fluctuations in the automatic bottling machine, the actual volume per bottle varies according to a normal distribution. It is observed that 10% of bottles are under 1.45L whereas 5% contain more than 1.55L. Calculate the mean and standard deviation of the volume distribution。


1
Expert's answer
2021-11-08T20:41:07-0500

Let X=X= the volume per bottle: XN(μ,σ2)X\sim N(\mu, \sigma^2)


P(X<1.45)=P(Z<1.45μσ)=0.1P(X<1.45)=P(Z<\dfrac{1.45-\mu}{\sigma})=0.1

1.45μσ1.28155\dfrac{1.45-\mu}{\sigma}\approx-1.28155


P(X>1.55)=1P(Z1.55μσ)=0.05P(X>1.55)=1-P(Z\leq\dfrac{1.55-\mu}{\sigma})=0.05

P(Z1.55μσ)=0.95P(Z\leq\dfrac{1.55-\mu}{\sigma})=0.95

1.55μσ1.64485\dfrac{1.55-\mu}{\sigma}\approx1.64485

1.45μ1.55μ=1.281551.64485\dfrac{1.45-\mu}{1.55-\mu}=\dfrac{-1.28155}{1.64485}

2.38503251.64485μ=1.9864025+1.28155μ2.3850325-1.64485\mu=-1.9864025+1.28155\mu

2.9264μ=4.3714352.9264\mu=4.371435

μ=1.493793\mu=1.493793

σ=1.451.4937931.2815\sigma=\dfrac{1.45-1.493793}{-1.2815}

σ=0.034173\sigma=0.034173

mean=μ=1.4938Lmean=\mu=1.4938L

σ=0.0342L\sigma=0.0342L


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