Question

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.

1. In a sample of six British citizens, what is the probability that two have ever boycotted goods for ethical reasons?
2. In a sample of six British citizens, what is the probability that at least two respondents have boycotted goods for ethical reasons?
3. In a sample of ten British citizens, what is the probability that between 3 and 6 have boycotted goods for ethical reasons?
4. 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

It is given that

Percentage of respondents who have boycotted goods for ethical reasons=23%. Let us call it p which is the probability of a British citizen who boycotted goods for ethical reasons:

There are only two possible outcomes of a British citizen ie eith boycott or not boycott.

p=0.23 q=1-p =0.77

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

Here n=6 , r=2 in a Binomial distribution.

Therfore 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?

Here again, n=6, p=0.23, q=0.77. Out of six, at least two have boycotted means 2 or more ie

  

SInce we know that the total probability is 1 and hence  

  

  

  

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

Here n=10. Between 3 and 6 have boycotted goods means x=3,4,5 and 6 and teh required probability is theerfore

=

  

  

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 here n=10, p=0.23 and 1-p=0.77. We also know that for a bionomial distribution, the mean=expected number of people that have boycotted goods =np

We also know that the variance

  

  

Also we know that