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sammy [17]
2 years ago
9

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)
Mathematics
1 answer:
NISA [10]2 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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