Question

In what follows, please round all probabilities to four decimal places. 1. The proportion of Canadian households that rely only on cellphones for communications instead of landline phones was estimated at 4.8% by a Statistics Canada residential telephone service survey. Suppose that 15 Canadian households are randomly selected. Let X be the number of Canadian households that rely only on cellphones for communications instead of landline phones among the selected 15 households. (a) (4 points) Find the probability that exactly 3 of the 15 Canadian households use only cellphones at home. (b) (4 points) Find the probability that at least one of the selected households uses only cellphones at home. (c) (4 points) Find the mean and standard deviation of X.

Answer 1

Here, X follows Binomial distribution

We have probability mass function for binomial distribution as :

P(X = x) = (.)p"(1 – p)-1

where,

n = Number of trials or sample size

p = Probability of success

x = Number of success

In our case,

n = 15

p = 4.8% = 0.048

(a)

Probability that exactly 3 of the 15 Canadian households use only cellphones at home = P(X=3)

P(X = 3) = ( 13 ) (0.048)*(1 – 0.048)25–

15! (0.048)(1-0.048) 12 3115_3ji.19301 -048"

Note:

n!= n*(n-1) *(n - 2)* ............. ska 1

P(X = 3) = 0.0279

(b)

Probability that at least one of the selected households uses only cellphones at home = P(X>0) = 1 - P(X=0)

P(X = 0) = () (0.048)º(1 – 0.048)25–

15! 10!(15 - 07(1 - 0.048)15

P(X=0) = 0.4781

Probability that at least one of the selected households uses only cellphones at home = 1 - 0.4781 = 0.5219

(c)

Mean = n*p = 15*0.048 = 0.72

Standard deviation = [ n*p*(1-p) ]^0.5 = [ 15*0.048*(1-0.048) ]^0.5 = 0.828