Question

Defects in computer hard-drives will usually render the entire computer worthless. For a particular model, the percent defective in the past has been 1%. If a sample size of 400 is taken, what would the 95.5% lower control chart limit be?

Answer 1

Answer:

0.00005

Step-by-step explanation:

We have the average fraction defective as p = 1%

= 0.01

N = 100

Then 95.5% deviation = 2σ

From there we have,

σ = √p*(1-p)/n

When we input these values, into the formula, we will have

σ = √0.01(1-0.01)/100

= √0.01(0.99)/100

= √0.000099

= 0.00995

The LCL = p - 2σ

= 0.01 - 0.00995

= 0.00005

lower control chart limit be is 0.00005

Answer 2

The 95.5% lower control chart limit be is 0.0005.

Given,

The average fraction percent in the past defective as p = 1%  = 0.01

Sample size N = 400

Then ,

According to the question ,

95.5% deviation = 2x

We have,

x = $\frac{\sqrt{p (1-p)} }{n}$

We put these values, into the formula,

x = $\frac{\sqrt{0.01 ( 1 - 0.01) } }{400}$

= $= \frac{\sqrt{(0.01) (0.99)} }{400}$

= $\frac{0.99}{400}$

= 0.0000245

The LCL = p - 2x

= 0.01 - 0.0000245

= 0.00005

The Lower control chart limit be is 0.00005

For the more information about Standard deviation click the link given below.

https://brainly.com/question/23907081