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3 years ago

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The weights of certain machine components are normally distributed with a mean of 8.04 g and a standard deviation of 0.08 g. Fin

d 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)

1 answer:

NISA [10]3 years ago

5 0

Answer:

The bottom 3 is separated by weight 7.8896 g and the top 3 is separated by weight 8.1904 g.

Step-by-step explanation:

We are given that

Mean, $\mu=8.04 g$

Standard deviation, $\sigma=0.08g$

We have to find the two weights that separate the top 3% and the bottom 3%.

Let x be the weight of machine components

$P(Xx_2)=0.03$

$P(X$

=0.03

From z- table we get

$P(Z1.88)=0.03$

Therefore, we get

$\frac{x_1-8.04}{0.08}=-1.88$

$x_1-8.04=-1.88\times 0.08$

$x_1=-1.88\times 0.08+8.04$

$x_1=7.8896$

$\frac{x_2-8.04}{0.08}=1.88$

$x_2=1.88\times 0.08+8.04$

$x_2=8.1904$

Hence, the bottom 3 is separated by weight 7.8896 g and the top 3 is separated by weight 8.1904 g.

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