Question

A manufacturer does not know the mean and standard deviation of the diameters for the production of ball bearings.However, a viewing system rejects all bearing larger than 2.4 cm and those under 1.8 cm in diameter.Out of 1000 ball bearings 8% are rejected as too small and 5.5% as too big.Find the mean and standard deviation of the ball bearings produced.

Answer 1

Answer on Question #44475-Physics-Other

A manufacturer does not know the mean and standard deviation of the diameters for the production of ball bearings. However, a viewing system rejects all bearing larger than $2.4\mathrm{cm}$ and those under $1.8\mathrm{cm}$ in diameter. Out of 1000 ball bearings $8\%$ are rejected as too small and $5.5\%$ as too big. Find the mean and standard deviation of the ball bearings produced.

Solution

Assume a normal distribution of

$$\Phi^ {- 1} (1 - 0. 0 8) = 1. 4;$$

so 1.8 is 1.4 standard deviations below mean.

Also

$$\Phi^ {- 1} (1 - 0. 0 5 5) = 1. 6,$$

so 2.4 is 1.6 standard deviations above mean.

This can be written as two simultaneous equations and solved:

$$\left\{ \begin{array}{l} \mu + 1. 6 \sigma = 2. 4 \\ \mu - 1. 4 \sigma = 1. 8. \end{array} \right.$$

Subtracting,

$$3. 0 \sigma = 0. 6 \rightarrow \sigma = 0. 2.$$

Using the first equation,

$$\mu + 1. 6 \cdot 0. 2 = 2. 4 \rightarrow \mu = 2. 0 8.$$

So, the mean is $\mu = 2.08$ and the standard deviation is $\sigma = 0.2$ .

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