According 10

According to the Board of Math Examiners, approximately 62% of people taking the Math Exam pass the exam.

If 18 people who have taken the Math Exam are randomly selected, what is the probability that at least 76% have passed? Round your answer to 4 decimal places.

If 46 people who have taken the Math Exam are randomly selected, what is the probability that at least 76% have passed? Round your answer to 4 decimal places.

Final Answer:

a = 0.1263
b = 0.0317

Step-by-step explanation:

p = 62 % = \frac{6 2}{1 0 0} = \frac{3 1}{5 0} = 0.62 r = 76 % = \frac{7 6}{1 0 0} = \frac{1 9}{2 5} = 0.76 n_{1} = 18 p_{1} = [r \cdot n_{1}] = [0.76 \cdot 18] = 14 n_{2} = 46 p_{2} = [r \cdot n_{2}] = [0.76 \cdot 46] = 35 p_{14} = (p _{1} n _{1}) \cdot p_{1}^{p} \cdot (1 - p)^{n_{1} - p_{1}} = (1 4 1 8) \cdot 0.6 2^{14} \cdot (1 - 0.62)^{18 - 14} = 3060 \cdot 0.6 2^{14} \cdot 0.3 8^{4} ≐ 0.0791 p_{15} = (p _{1} + 1 n _{1}) \cdot p_{1}^{p} + 1 \cdot (1 - p)^{n_{1} - p_{1} + 1} = (1 4 + 1 1 8) \cdot 0.6 2^{14} + 1 \cdot (1 - 0.62)^{18 - 14 + 1} ≐ 0.0344 p_{16} = (p _{1} + 2 n _{1}) \cdot p_{1}^{p} + 2 \cdot (1 - p)^{n_{1} - p_{1} + 2} = (1 4 + 2 1 8) \cdot 0.6 2^{14} + 2 \cdot (1 - 0.62)^{18 - 14 + 2} ≐ 0.0105 p_{17} = (p _{1} + 3 n _{1}) \cdot p_{1}^{p} + 3 \cdot (1 - p)^{n_{1} - p_{1} + 3} = (1 4 + 3 1 8) \cdot 0.6 2^{14} + 3 \cdot (1 - 0.62)^{18 - 14 + 3} ≐ 0.002 p_{18} = (p _{1} + 4 n _{1}) \cdot p_{1}^{p} + 4 \cdot (1 - p)^{n_{1} - p_{1} + 4} = (1 4 + 4 1 8) \cdot 0.6 2^{14} + 4 \cdot (1 - 0.62)^{18 - 14 + 4} ≐ 0.0002 a = p_{14} + p_{15} + p_{16} + p_{17} + p_{18} = 0.0791 + 0.0344 + 0.0105 + 0.002 + 0.0002 = 0.1263

p_{35} = (3 5 4 6) \cdot p^{35} \cdot (1 - p)^{46 - 35} = (3 5 4 6) \cdot 0.6 2^{35} \cdot (1 - 0.62)^{46 - 35} = 13340783196 \cdot 0.6 2^{35} \cdot 0.3 8^{11} ≐ 0.0172 p_{36} = (3 6 4 6) \cdot p^{36} \cdot (1 - p)^{46 - 36} = (3 6 4 6) \cdot 0.6 2^{36} \cdot (1 - 0.62)^{46 - 36} = 4076350421 \cdot 0.6 2^{36} \cdot 0.3 8^{10} ≐ 0.0086 p_{37} = (3 7 4 6) \cdot p^{37} \cdot (1 - p)^{46 - 37} = (3 7 4 6) \cdot 0.6 2^{37} \cdot (1 - 0.62)^{46 - 37} = 1101716330 \cdot 0.6 2^{37} \cdot 0.3 8^{9} ≐ 0.0038 p_{38} = (3 8 4 6) \cdot p^{38} \cdot (1 - p)^{46 - 38} = (3 8 4 6) \cdot 0.6 2^{38} \cdot (1 - 0.62)^{46 - 38} = 260932815 \cdot 0.6 2^{38} \cdot 0.3 8^{8} ≐ 0.0015 p_{39} = (3 9 4 6) \cdot p^{39} \cdot (1 - p)^{46 - 39} = (3 9 4 6) \cdot 0.6 2^{39} \cdot (1 - 0.62)^{46 - 39} = 53524680 \cdot 0.6 2^{39} \cdot 0.3 8^{7} ≐ 0.0005 p_{40} = (4 0 4 6) \cdot p^{40} \cdot (1 - p)^{46 - 40} = (4 0 4 6) \cdot 0.6 2^{40} \cdot (1 - 0.62)^{46 - 40} = 9366819 \cdot 0.6 2^{40} \cdot 0.3 8^{6} ≐ 0.0001 p_{41} = 0 p_{42} = 0 p_{43} = 0 p_{44} = 0 p_{45} = 0 p_{46} = 0 b = p_{35} + p_{36} + p_{37} + p_{38} + p_{39} + p_{40} + p_{41} + p_{42} + p_{43} + p_{44} + p_{45} + p_{46} = 0.0172 + 0.0086 + 0.0038 + 0.0015 + 0.0005 + 0.0001 + 0 + 0 + 0 + 0 + 0 + 0 = 0.0317

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You need to know the following knowledge to solve this word math problem:

statistics

binomial distribution

combinatorics

probability

arithmetic

rounding

basic operations and concepts

percentages

Grade of the word problem

high school

According 10

Final Answer:

Step-by-step explanation:

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