# Probability and Statistics

 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).
 A 2 B 0.45 C 0.18 D 0.25
GATE ME 2022 SET-1   Engineering Mathematics
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
 A $\sqrt{np},np(1-2p)$ B $\sqrt{np}, \sqrt{np(1-p)}$ C $np,np$ D $np,np(1-p)$
GATE ME 2021 SET-2   Engineering Mathematics
Question 2 Explanation:
Mean= np
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).
 A 3 B 4 C 6 D 8
GATE ME 2021 SET-1   Engineering Mathematics
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}
 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
 A 0.39 B 0.86 C 0.5 D 0.61
GATE ME 2021 SET-1   Engineering Mathematics
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,
$P = 1 – 0.606 = 0.39$
 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).
 A 12.25 B 58.36 C 36.67 D 18.62
GATE ME 2021 SET-1   Engineering Mathematics
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.

### 3 thoughts on “Probability and Statistics”

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

2. • 