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 |
Suppose the probability that a coin toss shows "head" is p, where 0\lt
p\lt 1. The coin is tossed repeatedly until the first "head" appears. The expected number of tosses required is
p/\left ( 1-p \right ) | |
\left ( 1-p \right )/p | |
1/p | |
1/p^{2} |
Question 4 Explanation:
F(H) = p, let x = Number of tosses
\begin{aligned} &\begin{array}{|c|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 \ldots \\ \hline P(x) & p & (1-p) p & (1-p)^{2} p & & \\ \hline \end{array}\\ &\begin{aligned} E(x) &=\sum x_{i} P(x)=p+2(1-p) p+3(p)(1-p)^{2}+\ldots \\ &=p\left[1+2(1-p)+3(1-p)^{2}+\ldots\right] \\ &=p[1-(1-p)]^{-2}=\frac{1}{p} \end{aligned} \end{aligned}
\begin{aligned} &\begin{array}{|c|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 \ldots \\ \hline P(x) & p & (1-p) p & (1-p)^{2} p & & \\ \hline \end{array}\\ &\begin{aligned} E(x) &=\sum x_{i} P(x)=p+2(1-p) p+3(p)(1-p)^{2}+\ldots \\ &=p\left[1+2(1-p)+3(1-p)^{2}+\ldots\right] \\ &=p[1-(1-p)]^{-2}=\frac{1}{p} \end{aligned} \end{aligned}
Question 5 |
Box contains the following three coins.
i. A fair coin with head on one face and tail on the other face.
ii. A coin with heads on both the faces.
iii. A coin with tails on both the faces.
A coin is picked randomly from the box and tossed. Out of the two remaining coins in the box, one coin is then picked randomly and tossed. If the first toss results in a head, the probability of getting a head in the second toss is
i. A fair coin with head on one face and tail on the other face.
ii. A coin with heads on both the faces.
iii. A coin with tails on both the faces.
A coin is picked randomly from the box and tossed. Out of the two remaining coins in the box, one coin is then picked randomly and tossed. If the first toss results in a head, the probability of getting a head in the second toss is
\frac{2}{5} | |
\frac{1}{3} | |
\frac{1}{2} | |
\frac{2}{3} |
Question 5 Explanation:
Let \quad P\left(H_{2}\right)= Probability of getting head in second toss
P\left(H_{1}\right)= Probability of getting head in first toss
\begin{aligned} P\left(\frac{H_{2}}{H_{1}}\right) &=\frac{P\left(H_{2} \cap H_{1}\right)}{P\left(H_{1}\right)} \\ P\left(H_{1}\right) &=\frac{1}{3} \times \frac{1}{2}+\frac{1}{3} \times 1+\frac{1}{3} \times 0=\frac{1}{2} \end{aligned}
To get head in second toss when head came in first toss, following cases can be made
1. Fair coin :
\begin{aligned} \text { Both head coin }=\frac{1}{3} \times \frac{1}{2} \times \frac{1}{2} \times 1=\frac{1}{12} \\ \text { Both tail coin }=\frac{1}{3} \times \frac{1}{2} \times \frac{1}{2} \times 0=0 \end{aligned}
2. Both head coin :
\begin{aligned} \text { Fair coin }&=\frac{1}{3} \times 1 \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{12}\\ \text{Both tail coin }&=0\\ \therefore \qquad P\left(H_{2} \cap H_{1}\right)&=\frac{1}{12}+\frac{1}{12}=\frac{1}{6} \\ \therefore \qquad P\left(H_{2} / H_{1}\right)&=\frac{1 / 6}{1 / 2}=\frac{1}{3} \end{aligned}
P\left(H_{1}\right)= Probability of getting head in first toss
\begin{aligned} P\left(\frac{H_{2}}{H_{1}}\right) &=\frac{P\left(H_{2} \cap H_{1}\right)}{P\left(H_{1}\right)} \\ P\left(H_{1}\right) &=\frac{1}{3} \times \frac{1}{2}+\frac{1}{3} \times 1+\frac{1}{3} \times 0=\frac{1}{2} \end{aligned}
To get head in second toss when head came in first toss, following cases can be made
1. Fair coin :
\begin{aligned} \text { Both head coin }=\frac{1}{3} \times \frac{1}{2} \times \frac{1}{2} \times 1=\frac{1}{12} \\ \text { Both tail coin }=\frac{1}{3} \times \frac{1}{2} \times \frac{1}{2} \times 0=0 \end{aligned}
2. Both head coin :
\begin{aligned} \text { Fair coin }&=\frac{1}{3} \times 1 \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{12}\\ \text{Both tail coin }&=0\\ \therefore \qquad P\left(H_{2} \cap H_{1}\right)&=\frac{1}{12}+\frac{1}{12}=\frac{1}{6} \\ \therefore \qquad P\left(H_{2} / H_{1}\right)&=\frac{1 / 6}{1 / 2}=\frac{1}{3} \end{aligned}
Question 6 |
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 6 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 7 |
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 8 |
Two continuous random variables X and Y are related as
Y=2X+3
Let \sigma ^{2}_{X} and \sigma ^{2}_{Y} denote the variances of X and Y, respectively. The variances are related as
Y=2X+3
Let \sigma ^{2}_{X} and \sigma ^{2}_{Y} denote the variances of X and Y, respectively. The variances are related as
\sigma ^{2}_{Y}=2 \sigma ^{2}_{X} | |
\sigma ^{2}_{Y}=4 \sigma ^{2}_{X} | |
\sigma ^{2}_{Y}=5 \sigma ^{2}_{X} | |
\sigma ^{2}_{Y}=25 \sigma ^{2}_{X} |
Question 8 Explanation:
\begin{aligned} Y &=2 X+3 \\ \operatorname{Var}[Y] &=E\left[(Y-\bar{Y})^{2}\right] \\ E[Y] &=\bar{Y}=2 \bar{X}+3 \\ \operatorname{Var}[Y] &=E\left[(2 X+3-2 \bar{X}-3)^{2}\right] \\ &=E\left[4(X-\bar{X})^{2}\right] \\ &=4 \cdot E\left[(X-\bar{X})^{2}\right] \\ \sigma_{Y}^{2} &=4 \cdot \sigma_{X}^{2} \end{aligned}
Question 9 |
The shape of the cumulative distribution function of Gaussian distribution is
Horizontal line | |
Straight line at 45 degree angle | |
Bell-shaped | |
S-shaped |
Question 9 Explanation:
PDF:f(x)=\frac{1}{\sigma \sqrt{2 \pi}}e^{-(x-\mu )^2/(2\sigma ^2)}
CDF:F(x)=\frac{1}{2}\left [ 1+eff\left ( \frac{x-\mu }{\sigma \sqrt{2}} \right ) \right ]
Question 10 |
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 10 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}
There are 10 questions to complete.
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