\[According{\text{ }}to{\text{ }}emperical{\text{ }}rule,{\text{ }}how{\text{ }}much{\text{ }}data{\text{ }}lies{\text{ }}between\,\mu - \sigma \,and\,\mu + \sigma \,?\]

• A) 95.44%
• B) 99.73%
• C) 75.66%
• D) 68.26%

{\text{For the Poisson distribution P(X = 1) = }}\frac{{{e^{ - 0.135}}{{0.135}^1}}}{{1!}}{\text{ the mean value is:}}

• A) 1
• B) -0.135
• C) 0.135
• D) {{e^{ - 0.135}}}

\[{\text{For the Poisson distribution P(X = 1) = }}\frac{{{e^{ - 2}}{2^1}}}{{1!}}{\text{ the mean value is:}}\]

• A) 2
• B) 0
• C) 3
• D) 1

{\text{In a normal distribution how much area lies between }}\mu \pm 2\sigma

• A) 95.45%
• B) 85%
• C) 75%
• D) 94%

{\text{The mean of Uniform distribution defined on the interval [a,b] is:}}

• A) \left( {a - b} \right)
• B) \frac{{a - b}}{2}
• C) \left( {a + b} \right)
• D) \frac{{a + b}}{2}

{\text{In a normal distribution how much area lies between }}\mu \pm \sigma

• A) 65%
• B) 75%
• C) 68.26%
• D) 80%

\[{\text{For the Poisson distribution P(X = 1) = }}\frac{{{e^{ - 0.135}}{{0.135}^1}}}{{1!}}{\text{ the mean value is:}}\]

• A) 0.135
• B) 1
• C) \[{{e^{ - 0.135}}}\]
• D) -0.135

{\text{In}}\,{\text{a}}\,{\text{binomial}}\,{\text{distribution,formula}}\,{\text{of}}\,{\text{calculating}}\,s\tan {\text{dard deviation is}}

• A) \sqrt p
• B) \sqrt {npq}
• C) \sqrt {pq}
• D) \sqrt {np}

\[According{\text{ }}to{\text{ }}emperical{\text{ }}rule,{\text{ }}how{\text{ }}much{\text{ }}data{\text{ }}lies{\text{ }}between\,\mu - 2\sigma \,and\,\mu + 2\sigma \,?\]

• A) 95.44%
• B) 99.73%
• C) 50%
• D) 68.26%

Let \[ f(X_i ,Y_i ) \] be the joint probability function of the two discrete random variables X and Y where i = 1,2,3,....,m and j = 1,2,3,....,n, then the marginal probability function of X is defined as:

• A) Both of the above
• B) None of the above
• C) \[ g(x_i ) = \sum\nolimits_j {f(X_i ,Y_i )} \]
• D) \[ g(x_i ) = \sum\nolimits_i {f(X_i ,Y_i )} \]

\[ {\text{The mean of the sampling distribution of }}\overline x _1 - \overline x _2 {\text{, denoted by }}\mu _{\overline x _1 - \overline x _2 } {\text{is equal to the difference between repective}} \]

• A) Population Means
• B) Population Proportions
• C) Sample Means
• D) Sample Proportions

{\text{In a normal distribution how much area lies between }}\mu \pm 3\sigma

• A) 68%
• B) 89%
• C) 70%
• D) 99.73%

Let f(X_i ,Y_i ) be the joint probability function of the two discrete random variables X and Y where i = 1,2,3,....,m and j = 1,2,3,....,n, then the marginal probability function of X is defined as:

• A) None of the above
• B) g(x_i ) = \sum\nolimits_i {f(X_i ,Y_i )}
• C) g(x_i ) = \sum\nolimits_j {f(X_i ,Y_i )}
• D) Both of the above

\[{\text{In a normal distribution how much area lies between }}\mu \pm 3\sigma \]

• A) 70%
• B) 99.73%
• C) 89%
• D) 68%

\[{\text{The mean of Uniform distribution defined on the interval [a,b] is:}}\]

• A) \[\frac{{a + b}}{2}\]
• B) \[\left( {a - b} \right)\]
• C) \[\frac{{a - b}}{2}\]
• D) \[\left( {a + b} \right)\]

\[ \begin{gathered} {\text{If X and Y are two discrete r}}{\text{.v's with joint probability function f (X}}_i ,Y_j ), \ {\text{then the conditional distribution of X given Y is:}} \\ \end{gathered} \] given by

• A) \[ f\left( {{\text{X}}_i |Y_j } \right){\text{ }} = f\left( {X_i ,{\text{ Y}}_j } \right){\text{ }}/h\left( {Y_j } \right) \]
• B) Both of the above
• C) None of the above
• D) \[ f\left( {{\text{X}}_i |Y_j } \right){\text{ }} = f\left( {X_i ,{\text{ Y}}_j } \right){\text{ }}/g\left( {X_i } \right) \]

According{\text{ }}to{\text{ }}emperical{\text{ }}rule,{\text{ }}how{\text{ }}much{\text{ }}data{\text{ }}lies{\text{ }}between\,\mu - \sigma \,and\,\mu + \sigma \,?

• A) 68.26%
• B) 75.66%
• C) 95.44%
• D) 99.73%

\[{\text{In a normal distribution how much area lies between }}\mu \pm 2\sigma \]

• A) 95.45%
• B) 94%
• C) 75%
• D) 85%

\[According{\text{ }}to{\text{ }}emperical{\text{ }}rule,{\text{ }}how{\text{ }}much{\text{ }}data{\text{ }}lies{\text{ }}between\,\mu - 3\sigma \,and\,\mu + 3\sigma \,?\]

• A) 95.44%
• B) 50%
• C) 99.73%
• D) 68.26%

A random experiment has five outcomes in its sample space {s1, s2, s3, s4, s5}. If P(s1)=0.2,P(s2)=0.3, P(s3)=0.1 and P(s4)=0.2 then P(s5)=?

• A) 0.2
• B) 1
• C) 0.8
• D) 0.5