Within what values will 95.44 percent of sample means of this process fall, if samples of n = 8 are taken and the process is in control (random)?

This question is incomplete, here is the complete question:

Specifications for a part for a DVD player state that the part should weigh between 25.2 and 26.2 ounces. The process that produces the parts has a mean of 25.7 ounces and a standard deviation of .25 ounce. The distribution of output is normal. Use Table-A

a) What percentage of parts will not meet the weight specs? (Round your "z" value and final answer to 2 decimal places.

b) Within what values will 95.44 percent of sample means of this process fall, if samples of n = 8 are taken and the process is in control (random)

a) What percentage of parts will not meet the weight specs = 4.56%

b) values within which 95.44 percent of sample means of this process falls are;

UCL = 25.88 ounces

LCL = 25.52 ounces

Step-by-step explanation:

Given that;

Mean u = 25.7 ounces

Std deviation = 0.25 ounces

a)

Z-score (Upper) = (X- u) / s = (26.2 - 25.7) / 0.25 = 2

Z-score (Lower) = ( 25.2-25.7 ) / 0.25 = -2

using the T - table

For Z = 2.0

the area in the tail of the curve to the right of the mean (upper) = 0.4772

therefore;

Number of defective = 0.5000 - 0.4772 = 0.0228

These are errors on one side of normal distribution.

To get the total error, we say

Total error = 2 × 0.0228

Total error = 0.0456 ≈ 4.56% ( 2 decimal place )

b)

given that;

n = 8,

standard deviation = 0.25 ounce

Standard deviation of X = Std deviation / √n

= 0.25 /√8 = 0.088

Now for 95.44% of confidence interval, Z = 2

UCL = Mean + Z × Standard deviation of X

= 25.7 + 2 × 0.088

= 25.88 ounces

LCL = Mean - Z × Standard deviation of X

= 25.7 - 2 × 0.088

= 25.52 ounces