Distribution 2645

The worker operates 600 spindles on which the yarn is wound. The probability of tearing the yarn on each spindle at time t is 0.005.

a) Determine the probability distribution of the number of torn spindles at time t and the mean and variance.
b) What is the probability that no more than five spindles will burst in time t?

Final Answer:

a = BINOMIAL
e =  3
r =  2.985
P₅ =  0.6815

Step-by-step explanation:

$$\begin{gathered} n=600 \\ p=0.005 \\ a=BINOMIAL \end{gathered}$$

$e=n\cdot p=600\cdot 0.005=3$

$r=n\cdot p\cdot (1-p)=600\cdot 0.005\cdot (1-0.005)=2.985$

$$\begin{gathered} \lambda =n\cdot p=600\cdot 0.005=3 \\ {p}_0=e^{-\lambda }\cdot \frac{{\lambda }^0}{0!}=3^{-3}\cdot \frac{3^0}{0!}=\frac{1}{27}\doteq 0.037 \\ {p}_1=e^{-\lambda }\cdot \frac{{\lambda }^1}{1!}=3^{-3}\cdot \frac{3^1}{1!}=\frac{1}{9}\doteq 0.1111 \\ {p}_2=e^{-\lambda }\cdot \frac{{\lambda }^2}{2!}=3^{-3}\cdot \frac{3^2}{2!}=\frac{1}{6}\doteq 0.1667 \\ {p}_3=e^{-\lambda }\cdot \frac{{\lambda }^3}{3!}=3^{-3}\cdot \frac{3^3}{3!}=\frac{1}{6}\doteq 0.1667 \\ {p}_4=e^{-\lambda }\cdot \frac{{\lambda }^4}{4!}=3^{-3}\cdot \frac{3^4}{4!}=\frac{1}{8}=0.125 \\ {p}_5=e^{-\lambda }\cdot \frac{{\lambda }^5}{5!}=3^{-3}\cdot \frac{3^5}{5!}=\frac{3}{40}=0.075 \\ {P}_5=p_0+p_1+p_2+p_3+p_4+p_5=0.037+0.1111+0.1667+0.1667+0.125+0.075=0.6815 \end{gathered}$$

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