A machine that produces a major part for an airplane engine is monitored closely.In the past, 6% of the parts produced would be defective.With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is

A) 70.
B) 69.
C) 135.
D) 136.

To find the sample size needed to estimate a proportion with a certain level of confidence and margin of error, we use the formula: $n=\frac{Z^2\cdot p\cdot (1-p)}{E^2}$ , where $Z$ is the Z-score corresponding to the confidence level (for 95% confidence, $Z=1.96$ ), $p$ is the proportion (in this case, 0.06), and $E$ is the margin of error (0.04). Plugging in the values: $n=\frac{1.96^2\cdot 0.06\cdot (1-0.06)}{0.04^2}$ , which calculates to approximately 135.8. Since sample size must be a whole number, we round up to the nearest whole number, which is 136.