consider taking a random sample size of 25 from a population in which 42% of the people have type A blood. what is the probability that the sample proportion with type A blood will be greater than 0.44 ? Use the normal approximation to the binomial with continuity correction.

I found the mean which is 25 * .42 = 10.5 , and the standard deviation = 6.09 but I don't know how to continue , any help



p = people with Blood A Type = .42

q = people with other blood type = .58

Mean = $np = 10.5$

S.D = $\sqrt{(npq)} = \sqrt{(25*.42*.58)} = 2.468$

Using Normal Approximation with continuity correction:

$$P(X>.44) = 1-P(X<=.44) = 1-P\left(\frac{(X-\mu)}{\sigma}\le\frac{ (.44*25 +.5 - 10.5)}{2.468}\right)$$

$$ = 1-0.657335 = 0.342665$$

  • $\begingroup$ You've mentioned that you're using a continuity correction in solving the problem but haven't really attempted it. I'm a bit confused about the need for continuity corrections in problems involving the distribution of the sample proportion. Is it necessary at all to use continuity corrections, seeing as the sample proportion is not discrete unlike the variable "number of people with blood type A"? Also, is the given sample size of 25 people large enough to assume that the sample proportion statistic is normally distributed? $\endgroup$ – Indula Jun 6 at 5:43
  • $\begingroup$ I have edited the solution. Sorry, it was not clear as I had omitted the important step of adding 0.5 for continuity correction in the penultimate step. The answer is the same it is just that I had not shown it explicitly. When you move from discrete to continuous approximation, you have to account for conitunity correction. That is the general rule of thumb. $\endgroup$ – Satish Ramanathan Jun 6 at 7:46

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