Expert: Scott A Wilson - 6/13/2009

Question
When i look at the question, my head totally blank. I'm stuck, i cant work out which is n which is p,is it binomial distribution or normal approximation. Please explain totally and as briefly possible on this question and steps by steps how to workout: a die is biased so that the probability of obtaining a six is 0.25. The die is thrown 200 times. a) find the probability of obtaining a six on the die i) more than 60 times, ii)less than 45 times, iii) between 40 & 55 times(inclusive). b) how many throws would be required if the probability of obtaining at least one six is greater than 0.9? Many Thanks

Answer
a) The number of times is 200; the chance of a 6 is 0.25.
i) the chance of more than 60 can be apprxoximated with the normal distribution.  The average is np and the variance is np(1-p).
The standard deviation is the squareroot of variance.

This means the average is 200*0.25 = 50 6's.
The variance is approximated by 200(1/4)(1 - 1/4) = 50*3/4 = 300/4.
The standard deviation, then, is
√(300/4) = √3√100/√4 = 10√3 / 2 = 5√3.

To get a n on the dice, when approximating with a normal, is the same as getting between n-0.5 and n+0.5 since the normal is a real number.

i) The odds of obtaining 60 or more in the binomial distribution can be approximated by the number over 59.5 in the normal.  Subtracting the average from this value, we get 59.5 - 50, which is 9.5 over the mean.  9.5/(4√3) is how many standard deviations above the average the lower limit is.  That is 1.30.  Use a normal table to look up how many are over 1.30 standard deviations away.

ii) The average number of 6's was 50.  The number 45 and below in the binomial distribution is equivalant to the number 45.5 and below in the normal distribution.  That can be changed into 4.5 below the average.  Divide by the standard deviation of 5√3 and the answer is how many standard deviations below the average that the value is.
Find out the chances of being this many standard deviations away from the average or more.

iii) To get from 40 to 55 in the binomial, that is equivalant to
40 - 0.5 to 55 + 0.5 in the normal distribution, or 39.5 to 55.5.
To standardize these value so they can be looked up, we need to know how far they are from the average and divide that by the number of standard deviations.

They are from 10.5 below the average to 5.5 above it.
Divide by the standard deviation and to determine how many standard deviations those two values are away.  Look up the number that are between the average and that many standard deviations for each value.  Add the two values gotten together.  


b) The chance that no 6 is rolled is 0.75^n where n is the number of rolls.  To make the chances of rolling at least one 6 0.9 or better,
that would be the chances of no 6 0.1 or less.  It is known that the chance of no six is 1 - 0.25 = 0.75 = 3/4.
If two rolls are done, the chances of no six are (3/4)^2 = 9/16;
for three rolls, this would be (3/4)^3 = 27/64 = 0.5625;
four rolls gives us (3/4)^4 = 81/256 = 0.421875;
five rolls is 243/1,024 = 0.237304688;  
six rolls is 729/4,096. = 0.177978516;
seven rolls is 2,187/16,356 = 0.133483887;  
eight rolls is 6,561/65,536 = 0.100112915, which is almost there; and
nine rolls is 19,683/262,144 = 0.075084686, which is small enough.

This means that nine rolls are required.