Question

5.) 45% of students rent their textbooks. You select a random group of 10 students. Find the probability that:a.) no more than 2 rent their textbookb.) at least one doesc.) find μ and σ

Answer 1

a) P = 0.0995

b) P = 0.9975

c) μ = 4.5

σ = 1.57

Step-by-step explanation:

To find the probabilities, we will use the binomial distribution because we have 10 identical events with a probability of 45% to success. Then, the probability that x students rent their textbook is calculated as

`\begin{gathered} P(x)=nCx\cdot p^x\cdot(1-p)^(n-x) \\ \text{ Where nCx = }(n!)/(x!(n-x)!) \end{gathered}`

Where n = 10 and p = 0.45, so

`P(x)=10Cx\cdot0.45^x\cdot(1-0.45)^(10-x)`

Then, the probability that no more than 2 rent their book is equal to

`\begin{gathered} P(x\leq2)=P(0)+P(1)+P(2) \\ \\ Where \\ P(0)=10C0\cdot0.45^0\cdot(1-0.45)^(10-0)=0.0025 \\ P(1)=10C1\cdot0.45^1\cdot(1-0.45)^(10-1)=0.0207 \\ P(2)=10C2\cdot0.45^2\cdot(1-0.45)^(10-2)=0.0763 \\ \\ Then \\ P(x\leq2)=0.0025+0.0207+0.0763 \\ P(x\leq2)=0.0995 \end{gathered}`

Now, we can calculate the probability that at least one does as

`\begin{gathered} P(x\ge1)=1-P(0) \\ P(x\ge1)=1-0.0025 \\ P(x\ge1)=0.9975 \end{gathered}`

Finally, the mean and standard deviation in a binomial distribution is equal to

`\begin{gathered} \mu=np \\ \mu=10(0.45) \\ \mu=4.5 \\ \\ \sigma=√(np(1-p)) \\ \sigma=√(10(0.45)(1-0.45)) \\ \sigma=1.57 \end{gathered}`

Therefore, the answers are

a) P = 0.0995

b) P = 0.9975

c) μ = 4.5

σ = 1.57