Question 1 |

Let a random variable X follow Poisson distribution
such that

Prob(X = 1) = Prob(X = 2).

The value of Prob(X = 3) is __________ (round off to 2 decimal places).

Prob(X = 1) = Prob(X = 2).

The value of Prob(X = 3) is __________ (round off to 2 decimal places).

2 | |

0.45 | |

0.18 | |

0.25 |

Question 1 Explanation:

\begin{aligned}
P(x=1)&=P(x=2)\\
\frac{e^{-\lambda }\lambda }{1!}&=\frac{e^{-\lambda }\lambda ^2}{2!}\\
\therefore \lambda &=2\\
P(x=3)&=\frac{e^{-2}2^3}{3!}=0.18
\end{aligned}

Question 2 |

The mean and variance, respectively, of a binomial distribution for n
independent trials with the probability of success as p, are

\sqrt{np},np(1-2p) | |

\sqrt{np}, \sqrt{np(1-p)} | |

np,np | |

np,np(1-p) |

Question 2 Explanation:

Mean= np

Variance = npq = np(1 - p)

Variance = npq = np(1 - p)

Question 3 |

Consider a single machine workstation to which jobs arrive according to a Poisson distribution with a mean arrival rate of 12 jobs/hour. The process time of the workstation is exponentially distributed with a mean of 4 minutes. The expected number of jobs at the workstation at any given point of time is ________ (round off to the nearest integer).

3 | |

4 | |

6 | |

8 |

Question 3 Explanation:

\lambda=12 per hour, \frac{1}{\mu}=4 \min / \mathrm{Jobs}, \mu=15 \mathrm{~Jobs} / \mathrm{hr}

\begin{aligned} \rho &=\frac{12}{15}=\frac{4}{5} \\ L_{s} &=\frac{\rho}{1-\rho}=\frac{\frac{4}{5}}{1-\frac{4}{5}}=\frac{\frac{4}{5}}{\frac{1}{5}} \\ \frac{4}{5} \times \frac{5}{1} &=4 \text { Jobs } \end{aligned}

\begin{aligned} \rho &=\frac{12}{15}=\frac{4}{5} \\ L_{s} &=\frac{\rho}{1-\rho}=\frac{\frac{4}{5}}{1-\frac{4}{5}}=\frac{\frac{4}{5}}{\frac{1}{5}} \\ \frac{4}{5} \times \frac{5}{1} &=4 \text { Jobs } \end{aligned}

Question 4 |

Customers arrive at a shop according to the Poisson distribution with a mean of 10 customers/hour. The manager notes that no customer arrives for the first 3 minutes after the shop opens. The probability that a customer arrives within the next 3 minutes is

0.39 | |

0.86 | |

0.5 | |

0.61 |

Question 4 Explanation:

10 Customers arrived in 1 hour, \lambda =10/hr

For 3 minutes, \lambda =\frac{1}{2}/hr

P(X=0)=\frac{e^{-\lambda }\lambda ^x}{x!}=\frac{e^{-\frac{1}{2}}\left ( \frac{1}{2} \right )^0}{0!}=0.606

The probability that a customer arrives within the next 3 minutes,

P = 1 – 0.606 = 0.39

For 3 minutes, \lambda =\frac{1}{2}/hr

P(X=0)=\frac{e^{-\lambda }\lambda ^x}{x!}=\frac{e^{-\frac{1}{2}}\left ( \frac{1}{2} \right )^0}{0!}=0.606

The probability that a customer arrives within the next 3 minutes,

P = 1 – 0.606 = 0.39

Question 5 |

Robot Ltd. wishes to maintain enough safety stock during the lead time period between starting a new production run and its completion such that the probability of satisfying the customer demand during the lead time period is 95%. The lead time period is 5 days and daily customer demand can be assumed to follow the Gaussian (normal) distribution with mean 50 units and a standard deviation of 10 units. Using \phi ^{-1}(0.95)=1.64, where \phi represents the cumulative distribution function of the standard normal random variable, the amount of safety stock that must be maintained by Robot Ltd. to achieve this demand fulfillment probability for the lead time period is _______ units (round off to two decimal places).

12.25 | |

58.36 | |

36.67 | |

18.62 |

Question 5 Explanation:

\begin{aligned} \text { Safety stock }&=Z \times \sigma \\ \sigma &=\sqrt{10^{2}+10^{2}+10^{2}+10^{2}+10^{2}} \\ &=10 \sqrt{5} \\ &=22.3606 \\ \text { Safety stock } &=1.64 \times 22.3606 \\ &=36.67 \end{aligned}

There are 5 questions to complete.

Answer for Q ME 2020 SET-2 should be:- (B) ( Normally distributed, only if X and Y are independent_)

Q no. 53 ans should be “d”…..

We have updated answer as (D).