Question

 A University found that 20% of its students withdraw without completing the first semester courses. Assume that 6 students have registered for the course this semester. To compute the following questions use binomial probability distribution.

a.      Compute the probability that 2 or fewer will withdraw?

b.      Compute the probability that exactly 4 will withdraw?

c.       Compute the expected number of withdrawals?

d.     What is the variance of the number of withdrawals?

e.     What is the standard deviation of the number of withdrawals?

Answer 1

a. $p = 0.2 \Rightarrow q = 1 - p = 0.8$

Then the probabilities that 0,1 or 2 will withdraw are respectively equal

${P_6}(0) = {q^6} = {0.8^6} = {\rm{0}}{\rm{.262144}}$

${P_6}(1) = C_6^1p{q^5} = 6 \cdot 0.2 \cdot {0.8^5} = {\rm{0}}{\rm{.393216}}$

${P_6}(2) = C_6^2{p^2}{q^4} = 15 \cdot {0.2^2} \cdot {0.8^4} = {\rm{0}}{\rm{.24576}}$

Then the required probability is

$P(X \le 2) = {P_6}(0) + {P_6}(1) + {P_6}(2) = 0.262144 + 0.393216 + 0.24576 = 0.90112$

Answer: $P(X \le 2) = 0.90112$

b. We have

${P_6}(4) = C_6^4{p^4}{q^2} = 15 \cdot {0.2^4} \cdot {0.8^2} = {\rm{0}}{\rm{.01536}}$

Answer: ${P_6}(4) = {\rm{0}}{\rm{.01536}}$

c. The expected number is equal to the expected value. Then

$M(x) = np = 6 \cdot 0.2 = 1.2$

Answer: $M(x) = 1.2$

d. The variance of the number of withdrawals is $D(x) = npq = 6 \cdot 0.2 \cdot 0.8 = {\rm{0}}{\rm{.96}}$.

Answer: 0.96

e. The standard deviation of the number of withdrawals is $\sigma (x) = \sqrt {D(x)} = \sqrt {0.96} \approx 0.98$.

Answer: $\sigma (x) \approx 0.98$