The probability that a part manufactured by a company will be defective is 0.05. If 15 such parts are selected randomly and inspected, then the probability that at least two parts will be defective is (round off to two decimal places).

The correct answer is 0.17.

Step-by-Step Explanation:

Let $X$ represent the number of defective parts in the sample of 15 parts.

Given parameters from the problem:
- Total number of parts inspected, $n=15$
- Probability of a part being defective, $p=0.05$

The variable $X$ follows a Binomial distribution:
$X\sim B(n,p)=B(15,0.05)$

We can approximate this using a Poisson distribution with parameter $\lambda$ where:
$\lambda =np=15\times 0.05=0.75$

Thus, $X$ is approximately Poisson distributed:
$X\sim P(\lambda =0.75)$

The probability mass function of a Poisson distribution is given by:
$P(X=x)=\frac{e^{-\lambda }{\lambda }^x}{x!}$

We need to find the probability that at least two parts will be defective, which is $P(X\ge 2)$: $P(X\ge 2)=1-P(X<2)=1-\left[P\right(X=0)+P(X=1\left)\right]$

Let us calculate the individual probabilities:
$P(X=0)=\frac{e^{-0.75}{0.75}^0}{0!}=e^{-0.75}$
$P(X=1)=\frac{e^{-0.75}{0.75}^1}{1!}=0.75e^{-0.75}$

Now, substitute these back into our expression:
$P(X\ge 2)=1-(e^{-0.75}+0.75e^{-0.75})=1-e^{-0.75}(1+0.75)=1-1.75e^{-0.75}$

Using the value of $e^{-0.75}\approx 0.47237$:
$P(X\ge 2)\approx 1-1.75\times 0.47237=1-0.82665=0.17335$

Rounding off to two decimal places, we get:
$P(X\ge 2)\approx 0.17$