Question #341256

manufacturer does not know the mean and SD of the diameters of ball bearings he is 

producing. However, a sieve system rejects all bearings larger than 2.4cm and those under 

1.8cm in diameter. Out of 1000 ball bearings 8% are rejected as too small and 5.5% as too 

big. What is the mean and standard deviation of the ball bearings produced assuming that 

the diameters of ball bearings are normally distributed?


1
Expert's answer
2022-05-16T15:37:57-0400
P(X<x1)=P(Z<x1μσ/n)P(X<x_1)=P(Z<\dfrac{x_1-\mu}{\sigma/\sqrt{n}})

=P(Z<1.8μσ/1000)=0.08=P(Z<\dfrac{1.8-\mu}{\sigma/\sqrt{1000}})=0.08

1.8μσ/1000=1.40507\dfrac{1.8-\mu}{\sigma/\sqrt{1000}}=-1.40507


P(X>x2)=1P(Xx2)P(X>x_2)=1-P(X\le x_2)

=1P(Zx2μσ/n)=1-P(Z\le\dfrac{x_2-\mu}{\sigma/\sqrt{n}})

=1P(Z2.4μσ/1000)=0.055=1-P(Z\le\dfrac{2.4-\mu}{\sigma/\sqrt{1000}})=0.055

2.4μσ/1000=1.59819\dfrac{2.4-\mu}{\sigma/\sqrt{1000}}=1.59819

2.4μ1.8μ=1.598191.40507\dfrac{2.4-\mu}{1.8-\mu}=\dfrac{1.59819}{-1.40507}

3.372168+1.40507μ=2.8767421.59819μ-3.372168+1.40507\mu=2.876742-1.59819\mu


μ=2.0807\mu=2.0807

2.42.0807σ/1000=1.59819\dfrac{2.4-2.0807}{\sigma/\sqrt{1000}}=1.59819

σ=6.318\sigma=6.318

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