Answer to Question #140748 in Statistics and Probability for Esther Gathenya

Question

A project company hired 800 casual employees to undertake a task for 1 week. On average (daily)
these workers take 10 hours to complete the task with a s.d deviation of 2 hours. If completion time is
normally distributed, determine the no. of workers who completed the task in:-
i. Less than 8 hours
ii. Less than 12.5 hours
iii. More than 13.5 hours
iv. Between 7.5 hours and 14 hours
v. Between 6.2 hours and 10 hours
vi. If the firm decides to lay off all the workers who cannot complete the task within hours, how
many workers will the firm lay-off?

Answer 1

Let $X=$ the daily time to complete the task: $X\sim N(\mu, \sigma^2).$

Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0, 1)$

Given $\mu=10 \ hours, \sigma=2 \ hours, N=800$

i.

$$P(X<8)=P(Z<\dfrac{8-10}{2})=P(Z<-1)\approx$$ $$\approx0.158655$$

$n=0.158655\times 800=125$

ii.

$$P(X<12.5)=P(Z<\dfrac{12.5-10}{2})=P(Z<1.25)\approx$$ $$\approx0.894350$$

$n=0.894350\times 800=715$

iii.

$$P(X>13.5)=1-P(X\leq13.5)=$$

$$=1-P(Z\leq\dfrac{13.5-10}{2})=P(Z\leq1.75)\approx$$

$$\approx1-0.959941=0.040059$$

$n=0.040059\times 800=32$

iv.

$$P(7.5<X<14)=P(X<14)-P(X\leq7.5)=$$

$$=P(Z<\dfrac{14-10}{2})-P(Z\leq\dfrac{7.5-10}{2})=$$

$$=P(Z<2)-P(Z\leq-1.25)\approx$$

$$\approx0.977250-0.105650=0.8716$$

$n=0.8716\times 800=697$

v.

$$P(6.2<X<10)=P(X<10)-P(X\leq6.2)=$$

$$=P(Z<\dfrac{10-10}{2})-P(Z\leq\dfrac{6.2-10}{2})=$$

$$=P(Z<0)-P(Z\leq-1.9)\approx$$

$$\approx0.5-0.028717=0.471283$$

$n=0.471283\times 800=377$

vi. Within 2 hours

$$P(10-2<X<10+2)=P(X<12)-P(X\leq8)=$$

$$=P(Z<\dfrac{12-10}{2})-P(Z\leq\dfrac{8-10}{2})=$$

$$=P(Z<1)-P(Z\leq-1)\approx$$

$$\approx1-2(0.158655)=0.682690$$

$n=800-0.682690\times 800=254$

The firm will lay-off 254 workers.