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Problem: In a binomial experiment with 45 trials, the probability of more than 25 successes can be approximated by $P(z > \frac{(25-27)}{3.29}$)

What is the probability of success of a single trial of this experiment?

Choices:

  • 0.07
  • 0.56
  • 0.79
  • 0.61
  • 0.6

My Solution: Since the experiment can be approximated by a normal distribution, I evaluated P(z) using NormalCdf on my calculator. $$P(z > \frac{25-27}{3.29}) = NormCdf(\frac{25-27}{3.29},\infty, 0, 1) = 0.72834 $$

Then, I set up the binomialCdf expression as $binomCdf(45, p, 26, 45)$ using 26 as the minimum number of successes as the question specifies it to be more than 25.

Through inspection, I found p to be 0.61. However, the answer key says otherwise.

Can someone check the error in my work? Thank you!

2 Answers 2

1

They're saying that \begin{align*} P(\text{Binom$(45,p)$} > 25) &\approx P(N(0,1) > \frac{25-27}{3.29}) \\ &= P(N(27,3.29^2) > 25) \end{align*} So the binomial is approximately equal in distribution to a $N(27,3.29^2)$ random variable. Setting the expected value in terms of the parameters of the binomial equal to the mean of the normal approximation, we have \begin{align*} E(\text{Binom$(45,p)$}) = np = 45 \cdot p \approx 27 \end{align*} So $p \approx 27/45 = 0.6$

1

When approximation a binomially distributed variable by means of the normal distribution one gets:

$$\mu = \frac{np}{\sigma} = \sqrt{np(1-p)}$$

Now $Z = \frac{X-\mu}{\sigma}$, so $\mu = np = 27$ and $\sigma = \sqrt{np(1-p)} = 3,29$.

Thus $\sigma = \sqrt{27(1-p)} = 3,29$ and solving yields $p=0,6$.