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If the expected time for completion of a project is 10 days with a standard deviation of 2 days, the expected time of completion of the project with 99.9% probability is

Correct Answer: 16 days

$$999\,\frac{1}{7} + 999\frac{2}{7} + 999\frac{3}{7} + $$     $$999\frac{4}{7} + $$   $$999\frac{5}{7} + $$   $$999\frac{6}{7}$$   is simplified to-

If the scheduled completion time of a project is more than the earliest expected time for completion of the project, then the probability of completion of the project within the scheduled completion time will be

Statement I The product of coefficient of correlation, standard deviation of variable 'X' and Standard deviation of variable 'Y' gives the measure of covariance between X and Y variables.
Statement II The product of coefficient of correlation between X and Y variables and the ratio between standard deviation of X variable to standard deviation of Y variable measures the slope of the regression line of X on Y variable.

A can complete a project in 20 days and B can complete the same project in 30 days. A and B start working on the project together and A quits 10 days before the project is expected to be completed. How many days in total will the project take to complete ?

If the expected time of completion of a project is 60 weeks with a standard deviation of 5 weeks, the probability of completing the project in 50 weeks and 65 weeks respectively will be

A mining project comprising of A, B and C activities is scheduled for 90 days at a cost of Rs. 1200 million. The manager of the project decides to reduce the time for completion of the project to 85 days. The decision was taken after 45 days, The minimum cost of the project in million rupees after crashing 5 days is
Mine System Engineering mcq question image

$${\text{999}}\frac{{998}}{{999}} \times 999$$   is equal to :

The standard normal cumulative probability function (probability from $$ - \infty $$  to xn) can be approximated as $${\text{F}}\left( {{{\text{x}}_{\text{n}}}} \right) = \frac{1}{{1 + {\text{exp}}\left( { - 1.7255{{\text{x}}_{\text{n}}}{{\left| {{{\text{x}}_{\text{n}}}} \right|}^{0.12}}} \right)}}$$       where xn = standard normal deviate. If mean and standard deviation of annual precipitation are 102 cm and 27 cm respectively, the probability that the annual precipitation will be between 90 cm and 102cm is

A class of first year B. Tech. students is composed of four batches A, B, C and D, each consisting of 30 students. It is found that the sessional marks of students in Engineering Drawing in batch Chave a mean of 6.6 and standard deviation of 2.3. The mean and standard deviation of the marks for the entire class are 5.5 and 4.2, respectively. It isdecided by the course instructor to normalize the marks of the students of all batches to have the same mean and standard deviation as that of the entire class. Due to this, the marks of a student in batch C are changed from 8.5 to

If, on an average, 'A' completes 40 tasks with a standard deviation of 5, 'B' completes 160 tasks with a standard deviation of 15 and 'C' completes 120 tasks with a standard deviation of 8, their order when arranged on the basis of consistency is