I'm doing some practical work on the binomial distribution but currently finding it difficult to answer iii-c. Here is the full question and the answers I've provided.
A bank claims that 80% of its customers use a cashpoint at least once a month. If this claim is true, what is the probability that:
i) in a random sample of 5 customers at least 80% use a cashpoint machine at least once a month
ii) in a random sample of 10 customers at least 80% use a cashpoint machine at at least once a month
iii-a) if the random variable X follows a binomial distribution with n=5 and p=0.8, what is the mean of this distribution and what is P(X $\ge$ mean)?
iii-b) if the random variable X follows a binomial distribution with n=10 and p=0.8, what is the mean of this distribution and what is P(X $\ge$ mean)?
iii-c) when the sample size n is increased, what happens to P(X $\ge$ mean)? If the sample size became very large what would you expect P(X $\ge$ mean) to approach? Briefly explain your answer.
These are the answers I have worked out so far:
i)
80% of 5=4
P (r=4) = ${}^5C_4$ * 0.8$^4$ * 0.2$^1$ = 0.4096
ii)
80% of 10=4
P(r=8) = ${}^{10}C_8$ * 0.8$^8$ * 0.2$^2$ = 0.3020
iii-a)
Mean = n * p = 5 * 0.8 = 4
P( r$\ge$4 )
= 1 - P(r<4)
= 1 - [ P(r=0) + P(r=1) +P(r=2) + P(r=3) ]
= 1 - 0.7235
= 0.2765
where
P(r=0) = ${}^{5}C_0$ * 0.8$^0$ * 0.2$^5$ = 0.0003
P(r=1) = ${}^{5}C_1$ * 0.8$^1$ * 0.2$^4$ = 0.0064
P(r=2) = ${}^{5}C_2$ * 0.8$^2$ * 0.2$^3$ = 0.0512
P(r=3) = ${}^{5}C_3$ * 0.8$^3$ * 0.2$^2$ = 0.2048
iii-b)
Mean = n * p = 10 * 0.8 = 4
P( r$\ge$8 )
= 1 - P(r<8)
= 1 - [ P(r=0) + P(r=1) +P(r=2) + P(r=3) + P(r=4) + P(r=5) + P(r=6) + P(r=7) ]
= 1 - 0.3223
= 0.6779
where
P(r=0) = ${}^{10}C_0$ * 0.8$^0$ * 0.2$^{10}$ = 0.0000
P(r=1) = ${}^{10}C_1$ * 0.8$^1$ * 0.2$^9$ = 0.0000
P(r=2) = ${}^{10}C_2$ * 0.8$^2$ * 0.2$^8$ = 0.0001
P(r=3) = ${}^{10}C_3$ * 0.8$^3$ * 0.2$^7$ = 0.0008
P(r=4) = ${}^{10}C_4$ * 0.8$^4$ * 0.2$^6$ = 0.0055
P(r=5) = ${}^{10}C_5$ * 0.8$^5$ * 0.2$^5$ = 0.0264
P(r=6) = ${}^{10}C_6$ * 0.8$^6$ * 0.2$^4$ = 0.0880
P(r=7) = ${}^{10}C_7$ * 0.8$^7$ * 0.2$^3$ = 0.2013
Thanks in advance!