Question

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

Answer 1

Answer:

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($\mu = 3, \sigma = 5^{2}$)

The z score probability distribution is given by;

            Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean value

            $\sigma$ = standard deviation

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

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

      P( $\frac{X-\mu}{\sigma}$ < $\frac{x-3}{5}$ ) = 0.95

      P(Z < $\frac{x-3}{5}$ ) = 0.95

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

                $\frac{x-3}{5}$ = 1.6449

                 $x -3 = 1.6449 \times 5$

                    $x$ = 3 + 8.22 = 11.22

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