It’s believed which 4% of young kids have a gene which might be related to youthful diabetes. Researchers during a organisation would identical to to exam brand new monitoring apparatus for diabetes. Hoping to have twenty young kids with a gene for their study, a researchers exam 732 incidentally comparison newborns for a participation of a gene related to diabetes.
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Dear mimmo,
You can use the Binomial distribution for calculating the probability of getting at least twenty children with the gene that may be linked to juvenile diabetes. Recall that with the binomial distribution, the probability of exactly k successes is given by
f(k) = C(n,k) p^k (1 – p)^(n – k),
for k in {0, 1, 2, . . . , n} (otherwise f(k) = 0),
where n is the number of trials, p is the probability of success on any particular trial, and C(n,k) = n! / [k! (n - k)!] (known as the binomial coefficient, which is the number of ways that exactly k items can be chosen from a set of n items).
For your question, n = 732 is the total number of newborns, k is the specific number of children with the suspect gene, and p = 0.04 is the probability of any particular child having that gene. With this we can compute the probability of exactly k children having the gene by calculating f(k). Note, however, that you want to know the probability of at least 20 children having the gene (since that will be enough subjects for the researchers’ study), which you can compute as follows.
P(at least 20 children with suspect gene)
= P(exactly 20 children with suspect gene) + P(exactly 21 children with suspect gene)
+ P(exactly 22 children with suspect gene) + . . . + P(exactly 732 children with suspect gene)
= f(20) + f(21) + f(22) + . . . + f(732).
Instead of computing all 713 terms, you can compute the first 20 terms and subtract them from 1:
P(at least 20 children with suspect gene)
= 1 – (f(0) + f(1) + f(2) + . . . + f(19)).
The values of f(0), f(1), f(2), . . . , and f(19) are obtained by substituting into the formula for f(k). For example,
f(19) = C(732,19) (0.04)^19 (1 – 0.04)^(732 – 19)
= 732! / [19! (732 - 19)!] (0.04)^19 (0.96)^713
= (732)(731)(730) . . . (714) / [(19)(18)(17) . . . (1)] (0.04)^19 (0.96)^713
= 0.010885 (to six decimal places).
The end result is that
P(at least 20 children with suspect gene)
= 1 – 0.026865
= 0.973135 (to six decimal places).
When either the number of terms or the magnitude of n grow large, the calculations using the binomial distribution can become cumbersome, especially when done manually. Computers are very helpful for these calculations, but not always available (such as on your final exam, perhaps). Approximations are frequently employed when exact computations are difficult to obtain, and the normal approximation to the binomial distribution is commonly used when n is large. The normal approximation also tends to work better when the binomial distribution is not very skewed (i.e., when p is closer to 0.5 rather than the extremes of 0 or 1).
For the normal approximation, start by taking the mean and variance of the binomial distribution and assign them to a normal distribution. Since the binomial distribution is discrete but the normal distribution is continuous, a correction is usually made by treating k from the binomial distribution as the interval for x between k – 0.5 and k + 0.5 with the normal distribution.
For the current problem, the mean is np = (732)(0.04) = 29.28 children and the variance is np(1 – p) = (732)(0.04)(0.96) = 28.1088 children^2 (giving a standard deviation of about 5.30 children). Thus, the normal approximation gives
P(at least 20 children with suspect gene)
= P(20 – 0.5 <= x <= 732 + 0.5)
= P(19.5 <= x <= 732.5)
= 1 – 0.032543
= 0.967457 (to six decimal places).
The normal approximation here matches the first two significant digits from the binomial distribution. This gives an absolute error of
|0.973135 – 0.967457| = 0.005678,
and a relative error of
|0.973135 – 0.967457| / 0.973135 = 0.005835,
or about 0.58%.