# (Random Sampling + Approximate Probability + Normal Approximation)

Suppose that 16% of the population of the U.S. is left-handed. If a random sample of 130 people from the U.S. is chosen, approximate the probability that at least 20 are left-handed. Use the normal approximation to the binomial with a correction for continuity. Round your answer to at least three decimal places. Do not round any intermediate steps.

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MY CALCULATIONS:

0.16 x 130 = 20.8

First you have to use the cdf of the binomial distribution:

$P(X \geq 20)=\sum_{x=20}^{130}{130 \choose x}\cdot \left(0.16\right)^x\cdot\left(0.84\right)^{130-x}$

Applying converse probability

$P(X \geq 20)=1-P(X \leq 19)=1-\sum_{x=0}^{19} {130 \choose x}\cdot \left(0.16\right)^x\cdot\left(0.84\right)^{1000-x}$

Approximation of the Binomial distribution by the Normal distribution:

$$P(X \geq 20)=1-\Phi\left(\frac{19+0.5-0.16\cdot 130}{\sqrt{130\cdot 0.16 \cdot 0.84}} \right)$$

0.5 is the continuity correction factor. And $\Phi(\cdot)$ is the cdf of the Standard Normal distribution. You can do the approximation, because $n\cdot p \cdot (1-p) >9$ (rule of thumb).

I think you can proceed from here.