Question

The weights of certain machine components are normally distributed with a mean of 8.14 g and a standard deviation of 0.09 g. Find the two weights that separate the top 3% and the bottom 3%. These weights could serve as limits used to identify which components should be rejected.

Answer 1

The weights of certain machine components are normally distributed with a mean of 8.14 g

Mean = 8.14g

Standard Deviation = 0.09g

The weight that weights that separate the top is express as: P(X>x);

The expression for the normal distribution is express as;

`P(Z=(x-\mu)/(\sigma))`

Since, it is given that distribution is 3%

so, P(Z) = 0.03

Substitute the value of Mean = 8.14g , P(Z) = 0.03and Standard Deviation = 0.09g

Now, for the bottom 3%; i.e. Z = 0

From the Z table P(0.03) = - 1.8808

`\begin{gathered} P(Z=(x-\mu)/(\sigma)) \\ 1.8808=(x-8.14)/(0.09) \\ x=7.88 \end{gathered}`

Now, for the distribution that seperate down, is 3%

From the Z table P(0.03) = 1.8808 since, it seperate down

So, P(Z) = - 1.8808

`\begin{gathered} P(Z=(x-\mu)/(\sigma)) \\ -1.8808=(x-8.14)/(0.09) \\ x=8.309 \end{gathered}`

Therefore, the limits that used to identify which components should rejected are 7.88 and 8.31

Answer : 7.88 and 8.31