Question

Problem 1 While performing scheduling work on a project you have determined that one activity has a great deal of variability. A sample of times (30) the task has been completed is provided the table below. Assume that the task is considered to be "late" if the task duration is two days greater than the mean value (rounded to the nearest whole day), and "early" if the task duration is two days less than the mean value (rounded to the nearest whole day). 24 28 26 15 17 25 27 29 16 19 19 17 23 24 18. 29 23 29 26 22 16 17 30 20 15 21 18 19 30 28 Assuming that the task duration follows a normal distribution, determine the likelihood: a) The task will take longer than 27 days b) Between 17 and 25 days c) The project will be early d) The project finishes during the 50% most likely late durations

Answer 1

Since the parameters of the normal distribution[
N(u, a2) N.c
], mean[$\small \mu$] and vanriance[$\small \sigma^2$] are not given in the question, we have to estimate them using the sample informations.

Now we know sample mean   
Σ/- i-1
is a very good estimator of population mean and the sample variance   
1 2 a - п) п — 12 i=1
is a very good estimator of population variance, where   $\small (x_1,x_2,.....,x_n)$    is a sample of size   $\small n$ .

Hence the estimated probability distribution for the task duration is   .

N(, s

From the given sample we get  

22.33333
   and  
225.12644
.

Hence the estimated probability distribution for the task duration is   $\small N(22.33333,25.12644)$ .

a.  

$\small P(X>27) =P\left ( \frac{X-22.33333}{5.012628}>0.930982 \right ) =1- \Phi(0.930982)$

= 1- 0.8240687 0.1759313

b.

$\small P(17<X<25) =P\left ( -1.063980<\frac{X-22.33333}{5.012628}<0.531990 \right )$

(0.531990) (-1.063980) 0.7026337 0.1436691 0.5589645

c. $\small P(\bar{x}-2 \geq X) =P\left (\frac{-2}{5.012628} \geq \frac{X-22.33333}{5.012628} \right ) =\Phi\left ( \frac{-2}{5.012628} \right )=0.3449494$

d.  

(I think there is some miss print in the question beacuse the question doesn't make since it says the chance of finishing the project late is 50%. And so there is probability to calculate. Hence assuming that I'm asked to find the probability of the project finishing late, i'm aswering the question.)

$\small P(\bar{x}+2 \leq X) =P\left (\frac{2}{5.012628} \leq \frac{X-22.33333}{5.012628} \right ) =1-\Phi\left ( \frac{2}{5.012628} \right )$

$\small =1-0.6550506=0.3449494$