Question

80. More lefties Refer to Exercise 72.

(a) Find the probability that exactly $3$students in the sample are left-handed. Show your work.
(b) Would you be surprised if the random sample contained $4$or more left-handed students? Compute $P(W\ge 4)$ and use this result to support your
answer.

Short answer

(a) The probability is $\mathrm{P}(\mathrm{X}=3)=12.85\%$

(b) The probability is $P(X\ge 4)\approx 5.56\%$and not surprised by $4$ or more left-handed students.

Step-by-step solution

Part (a) Step 1: Given information 

The probability that exactly $3$students in the sample are left-handed.

Part (a) Step 2: Explanation 

Given: $n=15$, and$p=0.10$

The Binomial Probability:

$\mathrm{P}(\mathrm{X}=\mathrm{k})=\left(\begin{array}{l}n\\ k\end{array}\right)\times p^k\times (1-p{)}^{n-k}$

Then for $k=3,$

$\mathrm{P}(\mathrm{X}=3)=\left(\begin{array}{l}15\\ 3\end{array}\right)\times (0.10{)}^3\times (0.90{)}^{15-3}$

localid="1650028682048" $$\begin{gathered} \approx 0.1285 \\ =12.85\% \end{gathered}$$.

Part (b) Step 1: Given information 

The random sample contained $4$ or more left-handed students. And compute for the value $P(W\ge 4)$.

Part (b) Step 2: Explanation 

Given: $n=20$and $p=0.85$

The Binomial Probability:

$\mathrm{P}(\mathrm{X}=\mathrm{k})=\left(\begin{array}{l}n\\ k\end{array}\right)\times p^k\times (1-p{)}^{n-k}$
$$\begin{gathered} \mathrm{P}(\mathrm{X}\ge 4)=\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)+\dots +\mathrm{P}(\mathrm{X}=15) \\ \approx 0.0556 \\ \approx 5.56\mathrm{\%} \end{gathered}$$