1 answer

A factory produces plastic cell phone cases. To fit properly, each case must have a width...

Question:

A factory produces plastic cell phone cases. To fit properly, each case must have a width between 53.5 and 54.5 millimeters. The quality control manager for the factory collects a random sample of 100 cases and determines that the widths are normally distributed, with a mean width of 54.2 millimeters and a standard deviation of 0.3 millimeter.

What percent of the cell phone cases meet manufacturing specifications?

Suppose the production line is adjusted so that the mean width is decreased to 54.0 millimeters and the standard deviation remains at 0.3 millimeter. What percent of cell phone cases will meet manufacturing specifications? In percents

Suppose that the mean width of the cell phone cases is 54.0 millimeters, and management would like 95% of the cases to meet manufacturing specifications. What standard deviation is required?


Answers

The z-score for X = 53.5 is

X-μ 53.5-54.2 ー-2.33 0.3

The z-score for X = 54.5 is

54.5-54.2 0.3 X-μ

The percent of the cell phone cases meet manufacturing specifications is

P(53.5 < X < 54.5)-P(-2.33 < < 1) 0.8314

Answer: 83.14%

----------------

The z-score for X = 53.5 is

53.5-54.0 0.3 X-μ

The z-score for X = 54.5 is

54.5-54.0 0.3 X-μ = 1.67

The percent of the cell phone cases meet manufacturing specifications is

P(53.5 < X < 54.5)-P(-1.67 < < 1.67) 0.9015

Answer: 90.15%

---------------------------------

Here we need z-scores that have 0.95 area between them. The z-scores +/- 1.96 have 0.95 area between them. So

z=rac{X-mu}{sigma}=rac{54.5-54.0}{sigma}

1.96=rac{54.5-54.0}{sigma}

sigma=0.2551

.

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