Question

It is claimed that 79% of all Americans live in cities with population greater than 100,000 people.Based on this, if 41 Americans are randomly selected, find the probability thata. Exactly 30 of them live in cities with population greater than 100,000 people.b. At most 33 of them live in cities with population greater than 100,000 people,c. At least 34 of them live in cities with population greater than 100,000 people,d, Between 28 and 33 (including 28 and 33) of them live in cities with population greater than100,000 people,HintoVideo on Finding Binomial Probabilities [+]Scratchwork Area

Answer 1

The Binomial probability distribution is given by:

$\begin{gathered} P(x)=n_{C_xP^x(1-P)^{n-x}}_{}_{} \\ \text{where:} \\ n\colon\text{ number of trials or }the\text{ number being sampled} \\ x\colon\text{ number of successes desired} \\ P\colon Probability\text{ of getting a success in one trial} \\ (1-P)\colon Probability\text{ of getting a failure in one trial} \end{gathered}$

From the question, we have the following:

$\begin{gathered} P=79\text{\%=0.79, I-P=1-0.79=0.21} \\ n=41 \end{gathered}$

a) Exactly 30 of them.

$\begin{gathered} P(30)=41_{C_{30}}(0.79)^{30}(0.21)^{41-30} \\ P(30)=0.094 \end{gathered}$

b) At most 33 people

$\begin{gathered} P(x\leq33)=1-P(x\ge34) \\ P(x\leq33)=1-\mleft\lbrace P(x=34\mright)+P(x=35)+P(x=36)+P(x=37)+P(x=38)+P(x=39)+P(x=40)+P(x=41)\} \end{gathered}$$\begin{gathered} P(x\leq33)=1-(0.1338+0.1007+0.0631+0.0321+0.0127+0.0036+0.00069+0.0000634 \\ P(x\leq33)=1-0.3467 \\ P(x\leq33)=0.6532 \end{gathered}$

c) At least 34 people

$P(x\ge34)=P(x=34)+P(x=35)+P(x=36)+P(x=37)+P(x=38)+P(x=39)+P(x=40)+P(x=41)$$\begin{gathered} P(x\ge34)=0.1338+0.1007+0.0631+0.0321+0.0127+0.0036+0.00069+0.0000634 \\ P(x\ge34)=0.3467 \end{gathered}$

d) Between 28 and 33 (including 28 and 33)

$\begin{gathered} P(28\leq x\leq33)=P(x=28)+P(x=29)+P(x=30)+P(x=31)+P(x=32)+P(x=33) \\ P(28\leq x\leq33)=0.0370+0.0624+0.0939+0.1253+0.1474+0.1512 \\ P(28\leq x\leq33)=0.6172 \end{gathered}$