Question

you are a consultant to a company that manufactures components for cell phones. one of the components the company manufactures has a 4% failure rate. design changes have improved the quality of the component. a test of 50 of the new components found that only one of the new components is defective.
a. before the design improvements, what was the probability that, among 50 of the items, at most one of the items was defective?
b. is it reasonable to conclude that the new components have a lower failure rate than 4%? explain.
c. would you recommend further testing to determine whether the new parts have a lower failure rate than 4%? explain.
a. the probability was about % (round to the nearest percent as needed.)

Explanation:

Step1: Identify the distribution

This is a binomial distribution problem. The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n = 50\), \(p=0.04\), and we want \(P(X\leq1)=P(X = 0)+P(X = 1)\)

Step2: Calculate \(P(X = 0)\)

The combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\). For \(k = 0\), \(C(50,0)=\frac{50!}{0!(50-0)!}=1\)
\(P(X = 0)=C(50,0)\times(0.04)^{0}\times(1 - 0.04)^{50-0}=1\times1\times0.96^{50}\)
Using a calculator, \(0.96^{50}\approx0.1299\)

Step3: Calculate \(P(X = 1)\)

For \(k = 1\), \(C(50,1)=\frac{50!}{1!(50 - 1)!}=\frac{50!}{1!49!}=50\)
\(P(X = 1)=C(50,1)\times(0.04)^{1}\times(1 - 0.04)^{50 - 1}=50\times0.04\times0.96^{49}\)
\(0.96^{49}\approx0.1353\), so \(P(X = 1)=50\times0.04\times0.1353 = 50\times0.005412=0.2706\)

Step4: Calculate \(P(X\leq1)\)

\(P(X\leq1)=P(X = 0)+P(X = 1)\approx0.1299+0.2706 = 0.4005\)
Converting to a percentage, \(0.4005\times100 = 40.05\%\approx40\%\)

Answer:

40