Question

The length of time (in seconds) an individual takes to use an ATM is Normally distributed with mean 100 and standard deviation 10. When Jackson arrives at an ATM, he finds 4 individuals waiting in a queue to use the machine. Let T be the duration (again, in seconds) of the period that starts when the first individual in queue accesses the ATM and ends when the last individual in queue leaves the ATM. Find P(T<360).

Answer 1

Answer:

0.1587

Step-by-step explanation:

let X be the length of time (in seconds) an individual takes to use an ATM

X is N(100,10)

T = the duration (again, in seconds) of the period that starts when the first individual in queue accesses the ATM and ends when the last individual in queue leaves the ATM.

Required probability = P(T<360)

Here T = 4x

Hence T is N with mean = 400 and std dev = 40

Or $\frac{T-400}{40}$ = Z is N(0,1)$P(T$

=0.5-0.3413

=0.1587