Question 1 |

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 1 Explanation:

Mean= np

Variance = npq = np(1 - p)

Variance = npq = np(1 - p)

Question 2 |

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 2 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 3 |

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 3 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 4 |

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 4 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}

Question 5 |

Consider a binomial random variable X. If X_1,X_2,..., X_n are independent and identically distributed samples from the distribution of X with sum Y=\sum_{i=1}^{n}X_i, then the distribution of Y as n\rightarrow \infty can be approximated as

Exponential | |

Bernoulli | |

Binomial | |

Normal |

Question 6 |

A fair coin is tossed 20 times. The probability that 'head' will appear exactly 4 times
in the first ten tosses, and 'tail' will appear exactly 4 times in the next ten tosses
is ______ (round off to 3 decimal places).

0.012 | |

0.052 | |

0.042 | |

0.068 |

Question 6 Explanation:

Fair coin tossed 20 times

First 10 times probability that head will appear exactly 4 times

P[4 heads in 10 tosses] ? P[4 tails in 10 tosses]

\begin{aligned} &=10_{C_{4}}\left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)^{6} \cdot 10_{C_{4}}\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{4} \\ &=\frac{10_{C_{4}} \cdot 10_{C_{4}}}{2^{10} \cdot 2^{10}}=\frac{210 \times 210}{2^{10} \times 2^{10}}=0.042 \end{aligned}

First 10 times probability that head will appear exactly 4 times

P[4 heads in 10 tosses] ? P[4 tails in 10 tosses]

\begin{aligned} &=10_{C_{4}}\left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)^{6} \cdot 10_{C_{4}}\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{4} \\ &=\frac{10_{C_{4}} \cdot 10_{C_{4}}}{2^{10} \cdot 2^{10}}=\frac{210 \times 210}{2^{10} \times 2^{10}}=0.042 \end{aligned}

Question 7 |

The sum of two normally distributed random variables X and Y is

always normally distributed | |

normally distributed, only if X and Y are independent | |

normally distributed, only if X and Y have the same standard deviation | |

normally distributed, only if X and Y have the same mean |

Question 7 Explanation:

\begin{aligned} X_{1} &\sim N\left(\mu_{1}, \sigma_{1}\right)\\ \text{and }\quad X_{2} &\sim N\left(\mu_{2}, \sigma_{2}\right)\\ \text{then }\quad X_{1}+X_{2} &\sim N\left(\mu_{1}+\mu_{2}, \sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}}\right) \end{aligned}

Always normally distributed.

Always normally distributed.

Question 8 |

Consider two exponentially distributed random variables X and Y, both having a mean
of 0.50. Let Z = X + Y and r be the correlation coefficient between X and Y. If the variance
of Z equals 0, then the value of r is ___________ (round off to 2 decimal places).

1 | |

-1 | |

-2 | |

2 |

Question 8 Explanation:

\begin{aligned} x \sim E\left(\lambda_{1}\right) ; \quad \text { mean } &=\frac{1}{\lambda_{1}}=0.5 \\ \Rightarrow \qquad \lambda_{1} &=2 \\ \text { Variance, } x &=\frac{1}{\lambda_{1}^{2}}=\frac{1}{4}=0.25 \\ Y \sim E\left(\lambda_{2}\right) ; \quad \text { Mean } &=\frac{1}{\lambda_{2}}=0.5 \\ \Rightarrow \quad \lambda_{2} &=2 \\ \text { Variance, } y &=0.25 \\ \text { Given Var }(\mathrm{Z})=\text{Var}(x)+& \text{Var}(\mathrm{y})+2 \mathrm{COV}(x, y) \\ 0 &=0.25+0.25+2 \mathrm{COV}(x, y) \\ COV(x,y)&=-\frac{0.5}{2}=-0.25\\ \text { Correlation, } \qquad \rho&=\frac{\text{COV}(x, y)}{\sigma_{x} \sigma_{y}} \\ &=\frac{-0.25}{\sqrt{(0.25) \sqrt{(0.25)}}}=-1 \end{aligned}

Question 9 |

A company is hiring to fill four managerial vacancies. The candidates are five men and
three women. If every candidate is equally likely to be chosen then the probability that
at least one women will be selected is _______ (round off to 2 decimal places).

0.21 | |

0.48 | |

0.93 | |

0.78 |

Question 9 Explanation:

5 men, 3 women

P[atleast one women selected for 4 vacancies]

\begin{array}{l} =1-P[\text { none }] \\ =1-\frac{5_{C_{4}} 3_{C_{0}}}{8_{C_{4}}}=1-\frac{5}{70}=1-\frac{1}{14}=\frac{13}{14}=0.93 \end{array}

P[atleast one women selected for 4 vacancies]

\begin{array}{l} =1-P[\text { none }] \\ =1-\frac{5_{C_{4}} 3_{C_{0}}}{8_{C_{4}}}=1-\frac{5}{70}=1-\frac{1}{14}=\frac{13}{14}=0.93 \end{array}

Question 10 |

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).

0.25 | |

0.17 | |

0.08 | |

0.55 |

Question 10 Explanation:

Let p= probability of making defective part

\begin{array}{c} =0.05 \\ \Rightarrow \mathrm{q}=1-\mathrm{p}=0.95 \end{array}

given n=15

Let X be number of defective parts be a random variable.

\begin{aligned} \mathrm{P}(\mathrm{X} \geq 2) &=1-\mathrm{P}(\mathrm{X} \lt 2) \\ &=1-\{\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)\} \\ &=1-\left(^{15} \mathrm{C}_{0} \mathrm{p}^{0} \mathrm{q}^{\mathrm{n}}+^{15} \mathrm{C}_{1} \mathrm{p}^{1} \mathrm{q}^{14}\right) \\ &\left.=1-\left\{(0.95)^{15}+15(0.05)(0.95)^{14}\right)\right\} \\ &=0.1709 \end{aligned}

\begin{array}{c} =0.05 \\ \Rightarrow \mathrm{q}=1-\mathrm{p}=0.95 \end{array}

given n=15

Let X be number of defective parts be a random variable.

\begin{aligned} \mathrm{P}(\mathrm{X} \geq 2) &=1-\mathrm{P}(\mathrm{X} \lt 2) \\ &=1-\{\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)\} \\ &=1-\left(^{15} \mathrm{C}_{0} \mathrm{p}^{0} \mathrm{q}^{\mathrm{n}}+^{15} \mathrm{C}_{1} \mathrm{p}^{1} \mathrm{q}^{14}\right) \\ &\left.=1-\left\{(0.95)^{15}+15(0.05)(0.95)^{14}\right)\right\} \\ &=0.1709 \end{aligned}

There are 10 questions to complete.

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