Question

An Internet survey estimates that, when given a choice between English and French,60% of the population prefers to study English. Three students are randomly selected and asked which of the two languages they prefer.

(a) Find the probability distribution for Y , the number of students in the sample who prefer English.

(b) What is the probability that exactly one of the three students  prefer English?

(c) What are the mean and standard deviation for Y ?

(d) What is the probability that the number of students prefer English falls within 2 standard deviations of the mean?

 

Answer 1

Solution

we have  \( P(English)=0.6 \)

(a) Find the probability distribution for Y 

Y : the number of students in the sample who prefer English.

Let : E=English , F=French , then \( Y\sim Bin(3,0.6) \)

Then \( P(Y=y)=C_3^y\left(0.6\right)^y\left(0.4\right)^{3-y} \)

\( \implies P(Y=0)=1\times (0.4)^3=0.064 \)

         \( P(Y=1)=3\times 0.6\times(0.4)^2=0.288 \)

         \( P(Y=2)=3\times (0.6)^2\times0.4=0.432 \)

         \( P(Y=3)=(0.6)^3=0.216 \)

(b) What is the probability that exactly one of the three  students prefer English

Exactly one \( \implies P(Y=1)=0.288 \)

(c) What are the mean and standard deviation for Y

Mean and  standard deviation for Y.

\( \implies E(Y)=np=3\times 0.6=1.8 \)

          \( V(Y)=npq=1.8\times 0.4=0.72 \)

           \( \delta(Y)=\sqrt{V(x)}=0.84 \)

(d) What is the probability that the number of students prefer English 

falls within \( 2\delta(Y) \) of the mean

\( \implies P(|y-E(Y)|<2\delta(Y) \)\( =P(E(Y)-2\delta(Y) \)\( < Y < E(\lambda)+2\delta(Y)) \)

Then, \( E(\lambda)-2\delta(Y)=0.12\hspace{2mm},E(\lambda)+2\delta(Y)=3.48 \)

\( \implies P(0.12< Y <3.48)=P(1\leq Y \leq 3)=0.936 \)

Therefore.

a).

b). \( P(Y=1)=0.288 \)

c). \( E(Y)=1.8 \),\( V(Y)=0.72 \),\( \delta(Y)=0.84 \)

d). \( P(|y-E(Y)|<2\delta(Y))=0.936 \)