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Virty [35]
4 years ago
12

A quality-control technician has been monitoring the output of a milling machine. Each day, the technician selects a random samp

le of 20 parts to measure and plot on the control chart. Over 10 days, the average diameter was 1.327 millimeters with a standard deviation of 0.0525 millimeters. What is the lower control limit (LCL) for an X-bar chart of this data? ________ millimeters
Business
1 answer:
aleksandrvk [35]4 years ago
6 0

Answer: 1.292 millimeters

Explanation:

Given the following ;

Number of Random sample (n) = 20

Average diameter(mean) = 1.327 millimeters

Standard deviation(std) = 0.0525 millimeters

To calculate the Lower Control Limit (LCL) :

Recall:

LCL = mean - [3(std÷sqrt(n))]

LCL = 1.327 - [3(0.0525÷sqrt(20))]

LCL = 1.327 - [3(0.0525 ÷ 4.472135955)]

LCL = 1.327 - [3 × 0.0117393568818]

LCL = 1.327 - 0.0352180706456216

LCL = 1.2917819293543784

LCL = 1.292 millimeters

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Quality services is an organization that operates several companies that market food products, restaurant equipment, and paper a
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Tina's aquarium has a capacity of 40 gallons of water she plans to stock her aquarium with 0.25 inches of fish per quart of Capa
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Answer:

The number of mollies will be 20

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7 0
4 years ago
Any time you walk into a store, you are in a market setting.
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5 0
3 years ago
Read 2 more answers
AB Builders, Inc., has 17-year bonds outstanding with a par value of $2,000 and a quoted price of 94.863. The bonds pay interest
timofeeve [1]

Answer:

13.47%

Explanation:

yield to maturity = {coupon + [(face value - market value)/n]} / [(face value + market value)/2]

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coupon + $3.0218 = $1,948.63 x 7.07% = $137.7681

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8 0
4 years ago
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