Question

In the US, 50.8% of the people are female. Suppose that 10 people in the US were chosen at random. Find the probability that fewer than 6 of those people are female.

Answer 1

We are given a binomial probability distribution problem.

Probability of success: p = 50.8% = 0.508

Number of trials: n = 10 people

Fewer than 6 means x = 0, 1, 2, 3, 4, 5

Recall that the binomial probability distribution is given by

`P(x)=^nC_x\cdot p^x\cdot(1-p)^(n-x)`

Where nCx is the number of possible combinations.

`P(x<6)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)`

For p(x = 0):

`\begin{gathered} P(x=0)=^(10)C_0\cdot0.508^0\cdot(1-0.508)^(10-0) \\ P(x=0)=0.00083 \end{gathered}`

For p(x = 1):

`\begin{gathered} P(x=1)=^(10)C_1\cdot0.508^1\cdot(1-0.508)^(10-1) \\ P(x=1)=10_{}\cdot0.508^1\cdot(0.492)^9 \\ P(x=1)=0.00858 \end{gathered}`

For p(x = 2):

`\begin{gathered} P(x=2)=^(10)C_2\cdot0.508^2\cdot(1-0.508)^(10-2) \\ P(x=2)=45\cdot0.508^2\cdot(0.492)^8 \\ P(x=2)=0.03987 \end{gathered}`

For p(x = 3):0

`\begin{gathered} P(x=3)=^(10)C_3\cdot0.508^3\cdot(1-0.508)^(10-3) \\ P(x=3)=120_{}\cdot0.508^3\cdot(0.492)^7 \\ P(x=3)=0.10978 \end{gathered}`

For p(x = 4):

`\begin{gathered} P(x=4)=^(10)C_4\cdot0.508^4\cdot(1-0.508)^(10-4) \\ P(x=4)=210\cdot0.508^4\cdot(0.492)^6 \\ P(x=4)=0.19836 \end{gathered}`

For p(x = 5):

`\begin{gathered} P(x=5)=^(10)C_5\cdot0.508^5\cdot(1-0.508)^(10-5) \\ P(x=5)=252\cdot0.508^5\cdot(0.492)^5 \\ P(x=5)=0.24578 \end{gathered}`

Finally, the probability is

`\begin{gathered} P(x<6)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5) \\ P(x<6)=0.00083+0.00858+0.03987+0.10978+0.19836+0.24578 \\ P(x<6)=0.6032_{}_{} \end{gathered}`

There is a 0.6032 (60.32%) probability that fewer than 6 of those people are female.