Question

Question 2

(a) It is noted that 8% of Kaplan students are left handed. If 20 (TWENTY) students are randomly

selected, calculate the

i. probability that none of them are left-handed, (2 marks)

ii. probability that at most 2 are left-handed, (3 marks)

iii. standard deviation for the number of left-handed students (2 marks)

(b) If 50 (FIFTY) classes of 20 (TWENTY) students are randomly selected, what is the probability that 10

(TEN) classes have no left-handed students?

Answer 1

Answer on Question #70020 – Math – Statistics and Probability

QUESTION

(a) It is noted that 8% of Kaplan students are left handed. If 20 (TWENTY) students are randomly selected, calculate the

i. probability that none of them are left-handed, (2 marks)

ii. probability that at most 2 are left-handed, (3 marks)

iii. standard deviation for the number of left-handed students (2 marks)

(b) If 50 (FIFTY) classes of 20 (TWENTY) students are randomly selected, what is the probability that 10 (TEN) classes have no left-handed students?

SOLUTION

(a)

i. $P(X = 0) = C_{20}^{0} 0.08^{0} 0.92^{20} = 0.92^{20} \approx 0.1887$.

ii. $P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = C_{20}^{0} 0.08^{0} 0.92^{20} + C_{20}^{1} 0.08^{1} 0.92^{19} + C_{20}^{2} 0.08^{2} 0.92^{18} = 0.7879$.

iii. $\sigma = \sqrt{np(1 - p)} = \sqrt{20 * 0.08 * 0.92} \approx 1.2133$.

(b)

$P(\text{no one in a class is left-handed}) = 0.92^{20} = 0.1887$

$P(\text{at least one in a class is left-handed}) = 1 - 0.1887 = 0.8113$

$P(10 \text{ classes have no left-handed students}) =$

$= C_{50}^{10} * 0.1887^{10} * 0.8113^{40} = 0.1370.$

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