Question

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?

Answer 1

$$P(X<x_1)=P(Z<\dfrac{x_1-\mu}{\sigma/\sqrt{n}})$$

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

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

$$P(X>x_2)=1-P(X\le x_2)$$

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

$$=1-P(Z\le\dfrac{2.4-\mu}{\sigma/\sqrt{1000}})=0.055$$

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

$$\dfrac{2.4-\mu}{1.8-\mu}=\dfrac{1.59819}{-1.40507}$$

$$-3.372168+1.40507\mu=2.876742-1.59819\mu$$

$$\mu=2.0807$$

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

$$\sigma=6.318$$