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

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

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 =
`(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(
`(X-\mu)/(\sigma)` <
`(x-3)/(5)` ) = 0.95

P(Z <
`(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;

`(x-3)/(5)` = 1.6449

`x -3 = 1.6449 * 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.