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Question

Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p.

Use the given values of n and p to find the mean μ and standard deviation σ.

Also, use the range rule of thumb to find the minimum usual value μ−2σ and the maximum usual value μ+2σ.

N = 1508, p = 3/4

 

 

PLEASE HELP WITH ANSWER

Answer 1

The binomial distribution can be solved exactly, but when dealing with large numbers of trials (large N), it's much simpler to use the normal approximation to it.

 

Under this approximation, the mean number of successes μ is given simply by N*p.  This should make sense - if you think of coin flips, for example, you expect to get heads half the time you flip the coin, so if you flip it a thousand times, you should get (on average) 1000 * .5 = 500 heads.  Here, we're performing 1508 trials with a success rate of 75%, so we expect a mean number of successes of 1131.

 

The standard deviation, meanwhile, can be calculated as √(N*p*(1-p)); here, √(1508*.75*.25) = √282.75 = 16.8.

 

Once we have these, it's simple to calculate μ−2σ = 1131-33.6 = 1097, and μ+2σ = 1131+33.6 = 1165.