- Why is the bar for k = 5 the tallest for 10 coin flips, but still only about 25 %?
- With n = 10 and p = 0.5 the mean is np = 5, so 5 heads is the single most likely count. There are C(10, 5) = 252 sequences with exactly five heads out of 2¹⁰ = 1024 equally likely sequences, giving P(K = 5) = 252/1024 ≈ 0.246. The remaining three quarters of the probability is spread over the other counts: 4 or 6 heads each have about 0.205, 3 or 7 about 0.117, and so on. 'Most likely' therefore does not mean 'likely'; the probability of landing within one standard deviation (√2.5 ≈ 1.58) of the mean, that is 4 ≤ K ≤ 6, is about 0.66.
- What does the continuity correction do, and why is it needed?
- The binomial variable takes only whole numbers, while a normal variable takes any real value. To approximate P(K ≤ k) with the normal curve, each integer k is represented by the interval from k − ½ to k + ½, so the whole event K ≤ k corresponds to the area under the curve up to k + ½, not up to k. Without the half the approximation systematically misses half a bar. For P(K = k) the corrected value is the area between k − ½ and k + ½, which is close to the height of the bar; with the correction the error for n = 20, p = 0.5 is below 0.001 at k = 10, while without it the normal gives P(K ≤ 10) ≈ 0.500 instead of the exact 0.588, an error of almost half a bar.
- When is the Poisson approximation good, and when should I not use it?
- The Poisson distribution with mean λ = np is the limit of the binomial as n grows and p shrinks with np held fixed. It is accurate when p is small, roughly p ≤ 0.05, and n is large enough that np is moderate; for n = 100, p = 0.03 the two distributions agree to within about 0.004 at every k. For p near 0.5 it is a poor approximation: the Poisson has variance λ = np, while the binomial has variance np(1 − p), which is only half as large at p = 0.5, so the Poisson stems are visibly too wide. The normal approximation is the right tool there.
- After 1000 experiments the pink outlines still do not match the cyan bars exactly. Is something wrong?
- No. Each pink bar is a relative frequency, a sample estimate of the corresponding exact probability, and it fluctuates from run to run. A bar whose true probability is q has standard error √(q(1 − q)/N) over N experiments; for q = 0.25 and N = 1000 that is about 0.014, so deviations of one or two hundredths are expected. As N increases the standard error shrinks like 1/√N: ten thousand experiments reduce it by a factor of about three, not ten. The same effect appears in the empirical mean, which approaches np but is not exactly np.