Answer to Question #248056 in Statistics and Probability for Agetro

Question

In a particular city, one in three families have a phone in their home. If 90 families are chosen at random, calculate the probability that at least 30 of them will have a phone.

Answer 1

Let $X=$ the number of families which have a phone: $X\sim Bin(n, p).$

Given $n=90, p=\dfrac{1}{3}, q=1-p=1-\dfrac{1}{3}=\dfrac{2}{3}.$

$$P(X\geq30)=1-P(X<30)$$

$$=1-\displaystyle\sum_{x=0}^{29}\dbinom{90}{x}(\dfrac{1}{3})^{x}(\dfrac{2}{3})^{90-x}$$

$$\approx0.53956753215$$ $$np=90(\dfrac{1}{3})=30\geq10, nq=90(\dfrac{2}{3})=60\geq10$$

Then $X$ has approximately a normal distribution with $\mu=np=30$ and $\sigma=\sqrt{npq}=\sqrt{90(\dfrac{1}{3})(\dfrac{2}{3})}=2\sqrt{5}.$

$$P(X\geq30)=P(X>29.5)=1-P(X\leq29.5)$$

$$=1-P(Z\leq\dfrac{29.5-30}{2\sqrt{5}})\approx1-P(Z\leq-0.1118)$$

$$\approx0.544509$$