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)
Answer:
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
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