Question

The number of complaints per day, X, received by a cable TV distributor has the probability distribution 2. 0 .4 .2 a) Find the expected number of complaints per day 0C6. 4)I(o.3)2 (0.1) 3( 0,2) b) Find the standard deviation of the number of complaints What is the approximate probability that the distributor receives a total of more than 100 complaints in 90 days? c)

Answer 1

c) The approximate probability that the distributor receives a total of more than 100 complaints in 90 days is 100 Since, on using excel function normsdist(z) 1-0.544236 → 0.455764 PX>100) 0.4558 Therefore, the approximate probability that the distributor receives a total of more than 100 complaints in 90 daysisP[X >100) 0.4558

Answer 2

SOLUTION :

a.

p = f(x) 

E(x) 

= ∑ f(x) * x  

= 0.4*0 + 0.3*1 + 0.2*2 + 0.1*3 

= 1.0  company / day (ANSWER).

b.

Variance 

= ∑ f(f)( x - E(x))^2

= 0.4(0 - 1)^2 + 0.3(1 - 1)^2 + 0.2(2 - 1)^2 + 0.1(3 - 1)^2

= 0.4 + 0 + 0.2 + 0.4

= 1.0 

So,

SD 

= sqrt(Variance) 

= sqrt(1.0)

= 1.0 company / day (ANSWER).

c.

Company per day = 100/90 = 10/9 

P(x > 10/9) 

= P(z > (10/9 - 1)/ 1)

= P(z > 0.1111)

From ND table 

= 1 - 0.5442

= 0.4558 (ANSWER).