Answer to Question #97619 in Statistics and Probability for Jericko

Question

You’re taking your kid sister trick or treating.If you know that an average house gives out 6 pieces of candy with a standard deviation 2.3. What is the probability that your kid sister will hit up a house giving out 9 pieces pf candy?

Answer 1

Let $X=$ the number of pieces of candy: $X\sim B(n,p).$

Then

$$mean=\mu_x=np$$ $$\sigma_x^2 =np(1-p)$$

$${\sigma_x^2 \over \mu_x}={np(1-p) \over np}=1-p$$

Given that $\mu_x=6, \sigma=2.3$

$$1-p={(2.3)^2 \over 6}={5.29 \over 6}$$ $$p=1-{5.29 \over 6}={0.71 \over 6}$$

$$n={\mu_x \over p}\approx51$$

$$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$

$$P(X=9)\approx\binom{51}{9}({0.71 \over 6})^9({5.29 \over 6})^{51-9}\approx0.0698$$

The probability that your kid sister will hit up a house giving out 9 pieces of candy is $0.0698$