A cell phone provider classifies its customers as Low users (less than 400 minutesper month) or High users (400 or more minutes per month). Studies have shownthat 80% of the people who were Low users one month will be Low users the nextmonth, and that 70% of the people who were High users one month will be Highusers the next month. (a) Set up the 2x2 stochastic matrix with columns and rows labeled L and H that displays these transitions
(b) Suppose that, during the month of January, 50% of the customers were Low users. What percent of the customers will be Low users in February? In March? (Type an integer or decimal for each matrix element.)

a ) See step by step explanation

b) 50%   and  50%

Step-by-step explanation:

The definition of type of user x is:

if x < 400 minutes  per month  low user

if x > 400 minutes per month  high user

The value for these states are of course yes  or no  ( the user is low or high user)

Then:

if the user is low today  xt  (0)    we have the probability of 80% ( 0,8) of have the user in the same condition and 20% (0,2) of having him or her as high user

xt = 1      

If the user is high today xt (1) is high today we have the probability of 70% ( 0,7) of having  the user in the same condition and 30% (0,3) of having him or her as high user

a) Then we construct the stochastic matrix

 Type of user

         (xt)               xt         xt+1  

      Low (0) or L   0,8        0,2

     High(1)  or H   0,3        0,7

The resulting equation system is:

xt  =  0,8* xt  +  0,2* xt+1

xt+1  = 0,3*xt + 0,7* xt+1

1  =  xt + xt+1

Solving for variables         xt+1  =  1  - xt

xt  = 0,8*xt + ( 1 - xt ) * 0,2

xt - 0,8*xt = 0,2 - 0,2xt

xt - 0,6*xt = 0,2

0,4*xt = 0,2

xt = 0,5       and     xt+1  = 0,5

b)  If in January 0,5  ( 50%) of customesr were  low user

In february  teh % of low user customer wll be

xt  =  0,8* xt  +  0,2* xt+1

0,5 = 0,8*0,5 + 0,2*xt+1

0,5 = 0,4 + 0,2 *xt+1

0,1/0,2 = xt+1        xt+1 = 0,5       or   50%

The same answer for March

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.