X is the time in hours required to find a bug in a software system. Assume that X is normally distributed with mean of 3 and standard deviation of 5. What is the lowest time limit in which 95% of the bugs can be found

The lowest time limit in which 95% of the bugs can be found is 11.22 hours.

Step-by-step explanation:

We are given that X is the time in hours required to find a bug in a software system. Assume that X is normally distributed with mean of 3 and standard deviation of 5.

So, X = time in hours required to find a bug in a software system

X ~ N()

The z score probability distribution is given by;

            Z = ~ N(0,1)

where, = mean value

            = standard deviation

Let the lowest time limit in which 95% of the bugs can be found =

Now, we have to find the lowest time limit in which 95% of the bugs can be found, i.e.;

      P( < ) = 0.95

      P(Z < ) = 0.95

Now, the critical value of in the z table which given an area of less than 95% is 1.6449, which means;

                = 1.6449

                    = 3 + 8.22 = 11.22

Therefore, the lowest time limit in which 95% of the bugs can be found is 11.22 hours.

the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation: