# Binomial distribution question

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

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Hint: Have you heard of the De Moivre-Laplace theorem? – Johannes Kloos Mar 14 '12 at 18:02
@JohannesKloos not at all - we haven't been taught that. What is it about? – methuselah Mar 14 '12 at 18:12
In (i) and (ii) you’ve calculated the probability that exactly 80% of the sample use a cashpoint machine at least once a month; what was wanted, however, was the probability that at least 80% do so. – Brian M. Scott Mar 14 '12 at 18:22
@BrianM.Scott Thanks a lot for pointing that out. I must have misread the question. This would make the answer for i) P(r<4) = 0.7235 and ii) P(r<8) = 0.3221 right? – methuselah Mar 14 '12 at 18:27

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.

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So the answers I proposed for i) P(r<4) = 0.7235 and ii) P(r<8) = 0.3221 are incorrect? – methuselah Mar 14 '12 at 18:57
Sorry it was just a few adding errors. I've got it now. Thanks for your time and patience! – methuselah Mar 14 '12 at 19:07
The value you got for $P(X<8)$ is correct except in the last decimal place. You were asked for $P(X\ge 8)$, and got the right answer, apart from roundoff error. In the first one, $P(X\ge 4)$, you set out the calculation for $P(X<4)$ correctly, did computations correctly. The sum however should be about $0.2623$. Then you should subtract from $1$. Except that it is more efficient to calculate probability of $4$, and of $5$, and add. The gain in efficiency is much greater in the case $n=10$. You calculated $8$ different numbers, added, subtracted from $1$. Need only $3$ numbers, then add. – André Nicolas Mar 14 '12 at 19:09
We weren't taught this rule in our lectures. So if P(X≥4) it is better to calculate the probability of 4 and 5 then add. Am I correct to think that if P(X≥8) I add up the probabilities of 8,9,10. What is the general rule for this? (i.e. in a case where P(X≥6) for example) – methuselah Mar 14 '12 at 19:26
It is not a question of rule, more like efficiency. For $P(X\ge 6)$, it doesn't make much difference. I would add together probabilities of $6$, $7$, $8$, $9$, $10$, so $5$ calculations plus addition. Your way is to add the results from $0$ to $5$, add, subtract from $1$. Yours is not much more work, $1$ more probability plus subtraction. But the difference is dramatic for $P(X\ge 9)$, the direct way is $2$ calculations plus addition, the indirect is $9$ calculations plus addition plus subtraction. – André Nicolas Mar 14 '12 at 19:37