Discussion 9

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9.1. BERNOULLI TRIALS pg. 337

10. Find the probability that among 10,000 random digits, the 3 appears no more than 931 times.

Area under the normal curve between Z=0 and Z= 2.28 is 0.4887 using the normal distribution table to two decimal points: https://www.mathsisfun.com/data/standard-normal-distribution-table.html

Digits: 0,1,2,3,4,5,6,7,8,9

\[\begin{aligned} &P\left(X_{10,000}>931\right)=P\left(X_{10,000} \geq 932\right) \\ &p: \frac{1}{10} \\ &\mu=n p=10,000 \times \frac{1}{10}=1,000 \\ &\sigma=\sqrt{n p(1-p)}=\sqrt{10,000 \times 0.1 \times 0.9}=30 \\ &z=\frac{931.5-10,000 \times 0.1}{\sqrt{10,000 \times 0.1 \times 0.9}}=-2.2833 \end{aligned}\]

Area under -2.28: 0.0113 Probability: 1.13%

(931.5-10000*.1)/sqrt(10000*.1*.9)

## [1] -2.283333

(1-2*0.4887)/2

## [1] 0.0113