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

Answer:

a) P = 0.0995

b) P = 0.9975

c) μ = 4.5

σ = 1.57

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

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

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

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

Therefore, the answers are

a) P = 0.0995

b) P = 0.9975

c) μ = 4.5

σ = 1.57