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16 days ago

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The weights of certain machine components are normally distributed with a mean of 8.01 g and a standard deviation of 0.06 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. Round to the nearest hundredth of a gram. 8.00 g and 8.02 g 7.88 g and 8.17 g 7.98 g and 8.04 g 7.90 g and 8.12 g

1 answer:

PIT_PIT [9.1K]16 days ago

7 0

Answer:

Option D) 7.90 g and 8.12 g

Step-by-step explanation:

The details provided in the question are:

Mean, μ = 8.01 g

Standard Deviation, σ = 0.06 g

The weights are distributed in a bell-shaped normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We need to determine the value of x for which the probability is 0.03.

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 8.01}{0.06})=0.03$

$= 1 -P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.03$

$=P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.97$

Using the standard normal z table, we find that,

$\displaystyle\frac{x - 8.01}{0.06} = 1.881\\\\x = 8.12$

Thus, the value of 8.17 g separates the upper 3% of the weights.

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 8.01}{0.06})=0.03$

From the standard normal z table, we derive,

$\displaystyle\frac{x - 8.01}{0.06} = -1.881\\\\x = 7.90$

Consequently, 7.90 separates the lower 3% of the weights.

<ptherefore the="" accurate="" answer="" is="">

Option D) 7.90 g and 8.12 g

</ptherefore>

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