Question

using excel software If 84.3% of Canadian homes have Internet access, find the probability that in a random sample of 15 Canadians that:

a. Exactly 13 have Internet access.

b. Between 12 and 14, inclusive, have Internet access.

c. At least 13 have Internet access.

d. At most 13 have Internet access.

Answer 1

BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is executed
p = success probability
mean = 15 * 0.843
= 12.645
II.
variance = npq
where
n = total number of repetitions experiment is executed
p = success probability
q = failure probability
variance = 15 * 0.843 * 0.157
= 1.9853
III.
standard deviation = sqrt( variance ) = sqrt(1.9853)
=1.409
a.
probability that exactly 13 have internet access
P( X = 13 ) = ( 15 13 ) * ( 0.843^13) * ( 1 - 0.843 )^2
= 0.281
b.
probability that between 12 and 14 inclusive have internet access
To find P(a < = Z < = b) = F(b) - F(a)
P( X < 12) = P(X=11) + P(X=10) + P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2)
= ( 15 11 ) * 0.843^11 * ( 1- 0.843 ) ^4 + ( 15 10 ) * 0.843^10 * ( 1- 0.843 ) ^5 + ( 15 9 ) * 0.843^9 * ( 1- 0.843 ) ^6 + ( 15 8 ) * 0.843^8 * ( 1- 0.843 ) ^7 + ( 15 7 ) * 0.843^7 * ( 1- 0.843 ) ^8 + ( 15 6 ) * 0.843^6 * ( 1- 0.843 ) ^9 + ( 15 5 ) * 0.843^5 * ( 1- 0.843 ) ^10 + ( 15 4 ) * 0.843^4 * ( 1- 0.843 ) ^11 + ( 15 3 ) * 0.843^3 * ( 1- 0.843 ) ^12 + ( 15 2 ) * 0.843^2 * ( 1- 0.843 ) ^13
= 0.1994
P( X < 14) = P(X=13) + P(X=12) + P(X=11) + P(X=10) + P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4)
= ( 15 13 ) * 0.843^13 * ( 1- 0.843 ) ^2 + ( 15 12 ) * 0.843^12 * ( 1- 0.843 ) ^3 + ( 15 11 ) * 0.843^11 * ( 1- 0.843 ) ^4 + ( 15 10 ) * 0.843^10 * ( 1- 0.843 ) ^5 + ( 15 9 ) * 0.843^9 * ( 1- 0.843 ) ^6 + ( 15 8 ) * 0.843^8 * ( 1- 0.843 ) ^7 + ( 15 7 ) * 0.843^7 * ( 1- 0.843 ) ^8 + ( 15 6 ) * 0.843^6 * ( 1- 0.843 ) ^9 + ( 15 5 ) * 0.843^5 * ( 1- 0.843 ) ^10 + ( 15 4 ) * 0.843^4 * ( 1- 0.843 ) ^11
= 0.7073
P(12 < X < 14) = 0.7073-0.1994 =0.5079
c.
probability that at least 13 have internet access
P( X < 13) = P(X=12) + P(X=11) + P(X=10) + P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) +
= ( 15 12 ) * 0.843^12 * ( 1- 0.843 ) ^3 + ( 15 11 ) * 0.843^11 * ( 1- 0.843 ) ^4 + ( 15 10 ) * 0.843^10 * ( 1- 0.843 ) ^5 + ( 15 9 ) * 0.843^9 * ( 1- 0.843 ) ^6 + ( 15 8 ) * 0.843^8 * ( 1- 0.843 ) ^7 + ( 15 7 ) * 0.843^7 * ( 1- 0.843 ) ^8 + ( 15 6 ) * 0.843^6 * ( 1- 0.843 ) ^9 + ( 15 5 ) * 0.843^5 * ( 1- 0.843 ) ^10 + ( 15 4 ) * 0.843^4 * ( 1- 0.843 ) ^11 + ( 15 3 ) * 0.843^3 * ( 1- 0.843 ) ^12
= 0.4262
P( X > = 13 ) = 1 - P( X < 13) = 0.5738
d.
probability that atmost 13 have internet access
P( X < = 13) = P(X=13) + P(X=12) + P(X=11) + P(X=10) + P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3)   
= ( 15 13 ) * 0.843^13 * ( 1- 0.843 ) ^2 + ( 15 12 ) * 0.843^12 * ( 1- 0.843 ) ^3 + ( 15 11 ) * 0.843^11 * ( 1- 0.843 ) ^4 + ( 15 10 ) * 0.843^10 * ( 1- 0.843 ) ^5 + ( 15 9 ) * 0.843^9 * ( 1- 0.843 ) ^6 + ( 15 8 ) * 0.843^8 * ( 1- 0.843 ) ^7 + ( 15 7 ) * 0.843^7 * ( 1- 0.843 ) ^8 + ( 15 6 ) * 0.843^6 * ( 1- 0.843 ) ^9 + ( 15 5 ) * 0.843^5 * ( 1- 0.843 ) ^10 + ( 15 4 ) * 0.843^4 * ( 1- 0.843 ) ^11 + ( 15 3 ) * 0.843^3 * ( 1- 0.843 ) ^12 +
= 0.7073