Question

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?

Answer 1

The question asks for the proportion of slots meeting width specifications within a normal distribution with defined mean and standard deviation. We calculate the corresponding Z-scores for the lower and upper specification limits and then determine the probability of a slot falling within these limits.

Step-by-step explanation:

The problem involves finding the proportion of slots that meet the specified width requirements in a normal distribution. In this case, the slot widths follow a normal distribution with a mean (μ) of 0.8750 inches, and a standard deviation (σ) of 0.0012 inches. The specifications require that slot widths be between 0.8725 inches and 0.8775 inches.

To find the proportion of slots that meet these specifications, we calculate the Z-scores for both the lower specification limit of 0.8725 and the upper specification limit of 0.8775. The Z-score formula is given by Z = (X - μ) / σ, where X is the value for which we want to find the Z-score.

For the lower limit, we have:

Z(lower) = (0.8725 - 0.8750) / 0.0012 = -2.083…

For the upper limit, we have:

Z(upper) = (0.8775 - 0.8750) / 0.0012 = 2.083…

Next, we use the standard normal distribution to find the probability corresponding to these Z-scores. The area under the curve between these two Z-scores represents the proportion of slots that are within the specifications. This can be found using standard normal distribution tables or a calculator with statistical functions.

Answer 2

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:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

P( widths between 0.8725 and 0.8775 inch)

`P(0.8725 \leq x \leq 0.8775) = P(\displaystyle(0.8725 - 0.8750)/(0.0012) \leq z \leq \displaystyle(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\%`

`P(0.8725 \leq x \leq 0.8775) = 96.2\%`

96.2% of slots meet these specifications.