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 monthii) 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!
Best Answer
There are errors of the same kind in i) and ii). In addition, the first two parts of iii) were done inefficiently, and there were a couple of numerical issues.
i) You were asked for the probability that at least $80$% in a random sample of $5$ use the machine. This is $$\binom{5}{4}(0.8)^4(0.2)^1 +\binom{5}{5}(0.8)^5(0.2)^0.$$
ii) The same type of mistake was made. We want the probability of exactly $8$ plus the probability of exactly $9$ plus the probability of exactly $10$.
iii-a) Same answer as for i). You did it the harder may, it is easier to add together probability of $4$, probability of $5$.
iii-b) Same answer as for ii). Again, you did it in much too hard a way.
iii-c) The distribution approaches a normal distribution ("bell-shaped curve") which is symmetric about the mean. So the probability that $X$ is greater than or equal to the mean approaches $1/2$.
Comments on computation: For iii-a), you accidentally wrote down the wrong probability. I get $0.73728$. Two errors here, you meant to subtract from $1$, you did, but then the two numbers were somehow transposed. There is also an error probably due to rounding. There is also a typo in which you wrote down the wrong exponent, but did the calculation right. For iii-b), and therefore for ii), I get $0.6777995$, which differs from your answer in the fourth decimal place. In the real world, no bih deal, the numbers like $80$\% are presumably not exact. But you should use the full capacities of your calculator, in particular the memory feature, to avoid rounding errors that add up. It will also save you time, and maybe cut down on mistakes. Rekeying takes time, and keying errors are easy to make.