Question

5. A survey on British Social Attitudes asked respondents if they had ever boycotted goods for ethical reasons (Statesman, January 28, 2008). The survey found that 23% of the respondents have boycotted goods for ethical reasons.

a) In a sample of six British citizens, what is the probability that two have ever boycotted goods for ethical reasons? .

b) In a sample of six British citizens, what is the probability that at least two respondents have boycotted goods for ethical reasons?

c) In a sample of ten British citizens, what is the probability that between 3 and 6 have boycotted goods for ethical reasons?

d) In a sample of ten British citizens, what is the expected number of people that have boycotted goods for ethical reasons? Also find the standard deviation.

Answer 1

a) In a sample of six British citizens, what is the probability that two have ever boycotted goods for ethical reasons?

Solution:

We have n = 6, p = 0.23, q = 1 – p = 1 – 0.23 = 0.77

We have to find P(X=2)

P(X=x) = nCx*p^x*q^(n – x)

P(X=2) = 6C2*0.23^2*0.77^(6 – 2)

P(X=2) = 15*0.23^2*0.77^4

P(X=2) = 0.278939

Required probability = 0.278939

b) In a sample of six British citizens, what is the probability that at least two respondents have boycotted goods for ethical reasons?

Solution:

We have n = 6, p = 0.23, q = 0.77

We have to find P(X≥2)

P(X≥2) = 1 – P(X<2) = 1 – P(X≤1)

P(X≤1) = P(X=0) + P(X=1)

P(X=x) = nCx*p^x*q^(n – x)

P(X=0) = 6C0*0.23^0*0.77^(6 – 0)

P(X=0) = 1*1*0.77^6

P(X=0) = 0.208422

P(X=1) = 6C1*0.23^1*0.77^(6 – 1)

P(X=1) = 6*0.23*0.77^5

P(X=1) = 0.373536

P(X≤1) = P(X=0) + P(X=1)

P(X≤1) = 0.208422+ 0.373536

P(X≤1) = 0.581958

P(X≥2) = 1 – P(X<2) = 1 – P(X≤1)

P(X≥2) = 1 – P(X<2) = 1 – 0.581958

P(X≥2) = 0.418042

Required probability = 0.418042

c) In a sample of ten British citizens, what is the probability that between 3 and 6 have boycotted goods for ethical reasons?

We have n = 10, p = 0.23, q = 0.77

We have to find P(3<X<6)

P(3<X<6) = P(X=4) + P(X=5)

P(X=x) = nCx*p^x*q^(n – x)

P(X=4) = 10C4*0.23^4*0.77^(6 – 4)

P(X=4) = 210*0.23^4*0.77^2

P(X=4) = 0.348427

P(X=5) = 10C5*0.23^5*0.77^(6 – 5)

P(X=5) = 252*0.23^5*0.77^1

P(X=5) = 0.124891

P(3<X<6) = P(X=4) + P(X=5)

P(3<X<6) = 0.348427 + 0.124891

P(3<X<6) = 0.473318

Required probability = 0.473318

d) In a sample of ten British citizens, what is the expected number of people that have boycotted goods for ethical reasons? Also find the standard deviation.

We have n = 10, p = 0.23, q = 0.77

Expected number = n*p = 10*0.23 = 2.3

Standard deviation = sqrt(npq) = sqrt(10*0.23*0.77) = 1.330789