Respuesta :
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([tex]\mu = 3, \sigma = 5^{2}[/tex])
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean value
[tex]\sigma[/tex] = standard deviation
Let the lowest time limit in which 95% of the bugs can be found = [tex]x[/tex]
Now, we have to find the lowest time limit in which 95% of the bugs can be found, i.e.;
P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{x-3}{5}[/tex] ) = 0.95
P(Z < [tex]\frac{x-3}{5}[/tex] ) = 0.95
Now, the critical value of [tex]x[/tex] in the z table which given an area of less than 95% is 1.6449, which means;
[tex]\frac{x-3}{5}[/tex] = 1.6449
[tex]x -3 = 1.6449 \times 5[/tex]
[tex]x[/tex] = 3 + 8.22 = 11.22
Therefore, the lowest time limit in which 95% of the bugs can be found is 11.22 hours.
