Question

Historical data indicates that only 20% of cable customers are willing to switch companies. If a binomial process is assumed, then in a sample of 20 cable customers, what is the probability that.
a. Between 2 and 5 (inclusive) customers are willing to switch companies?
b. Less that 5 customers are willing to switch.
c. More than 16 customers are willing to switch.

Answer 1

Answer:

a. The probability is 0.735

b. The probability is 0.6296

c. The probability is 0

Step-by-step explanation:

If we assume a binomial process, the probability that x customer are willing to switch companies is:

$P(x)=nCx*p^{x}*(1-p)^{n-x}$

nCx is calculated as:

$nCx=\frac{n!}{x!(n-x)!}$

Where n are the 20 cable customers and p is the probability 0.2 that the cable customer are willing to switch companies.

Then, P(x) is:

$P(x)=20Cx*0.2^{x}*(1-0.2)^{20-x}$

The probability that between 2 and 5 (inclusive) customers are willing to switch companies is:

P(2≤x≤5) = P(2) + P(3) + P(4) + P(5)

Where P(2), P(3), P(4) and P(5) are equal to:

$P(2)=20C2*0.2^{2}*(1-0.2)^{20-2}=0.1369$

$P(3)=20C3*0.2^{3}*(1-0.2)^{20-3}=0.2054$

$P(4)=20C4*0.2^{4}*(1-0.2)^{20-4}=0.2182$

$P(5)=20C5*0.2^{5}*(1-0.2)^{20-5}=0.1745$

So, P(2≤x≤5) is:

P(2≤x≤5) = 0.1369 + 0.2054 + 0.2182 + 0.1745 = 0.735

At the same way, the probability that less than 5 customers are willing to switch is:

P(x<5)=P(0)+P(1)+P(2)+P(3)+P(4)

P(x<5)=0.6296

Finally, the probability that more than 16 customers are willing to switch is:

P(x>16)=P(17)+P(18)+P(19)+P(20)

P(x>16)=0