Let maximum marks in the examination be \(100x\) and passing marks be \(y\)
Peter got 30% of the maximum marks, =\(30x+10=y\)-----(i)
Similarly, for Paul =\(40x-15=y\)----(ii)
Subtracting equation (i) from (ii) =\(10x=25\)
\(x=2.5\)
Substituting it in equation (ii), we get : Passing marks =\(y=85\)
The correct answer is (D) :85
List-I | List-II |
---|---|
(A) Confidence level | (I) Percentage of all possible samples that can be expected to include the true population parameter |
(B) Significance level | (III) The probability of making a wrong decision when the null hypothesis is true |
(C) Confidence interval | (II) Range that could be expected to contain the population parameter of interest |
(D) Standard error | (IV) The standard deviation of the sampling distribution of a statistic |