2. a factory is producing a machine part. the manager of the factory states that the probability of…

Explanation:

Step1: Identify distribution

This is a binomial distribution problem, where $n = 100$ (number of trials - number of parts produced), $p=0.03$ (probability of a part being defective).

Step2: Calculate probabilities

The probability mass function of a binomial distribution is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$. We want to find $P(X\geq4)=1 - P(X\lt4)=1-(P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3))$. For $k = 0$: $C(100,0)=\frac{100!}{0!(100 - 0)!}=1$, $P(X = 0)=1\times(0.03)^{0}\times(0.97)^{100}\approx0.0476$. For $k = 1$: $C(100,1)=\frac{100!}{1!(100 - 1)!}=100$, $P(X = 1)=100\times0.03\times(0.97)^{99}\approx0.1471$. For $k = 2$: $C(100,2)=\frac{100!}{2!(100 - 2)!}=\frac{100\times99}{2\times1}=4950$, $P(X = 2)=4950\times(0.03)^{2}\times(0.97)^{98}\approx0.2252$. For $k = 3$: $C(100,3)=\frac{100!}{3!(100 - 3)!}=\frac{100\times99\times98}{3\times2\times1}=161700$, $P(X = 3)=161700\times(0.03)^{3}\times(0.97)^{97}\approx0.2275$.

Step3: Calculate $P(X\geq4)$

$P(X\lt4)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)\approx0.0476 + 0.1471+0.2252 + 0.2275=0.6474$. $P(X\geq4)=1 - 0.6474 = 0.3526$.

Since $P(X\geq4)=0.3526\gt0$, it is possible to have 4 or more defective parts. So the manager is incorrect.

Answer:

The manager is incorrect because the probability of having 4 or more defective parts is $0.3526$, which means there is a non - zero chance of having 4 or more defective parts.