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Automated manufacturing operations are quite precise but still vary, often with distributions that are close to Normal. The width in inches of slots cut by a milling machine follows approximately the N(0.8750, 0.0012) distribution. The specifications allowslot widths between 0.8725 and 0.8775 inch. What proportion of slots meet these specifications?

Respuesta :

Answer:

96.2% of slots meet these specifications.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 0.8750

Standard Deviation, σ = 0.0012

We are given that the distribution of width in inches of slots is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

P( widths between 0.8725 and 0.8775 inch)

[tex]P(0.8725 \leq x \leq 0.8775) = P(\displaystyle\frac{0.8725 - 0.8750}{0.0012} \leq z \leq \displaystyle\frac{0.8775-0.8750}{0.0012}) = P(-2.083 \leq z \leq 2.083)\\\\= P(z \leq 2.083) - P(z < -2.083)\\= 0.981 - 0.019 = 0.962 = 96.2\%[/tex]

[tex]P(0.8725 \leq x \leq 0.8775) = 96.2\%[/tex]

96.2% of slots meet these specifications.

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