Answer:
a) The probability that exactly 8 said it was consolidating or reducing errands is 0.0815
b) The probability that none of them said it was cutting recreational driving is 0.0115
c) The probability that more than 7 said it was cutting recreational driving is 0.0321
Explanation:
The images attached show the step by step solution to solving the problem.
The problem is a binomial distribution problem with n = 20 and has different value for p.
(a): The probability that exactly 8 is [tex]0.0815[/tex]
(b): the probability that none of them [tex]0.144[/tex]
(c): The probability that more than 7 is [tex]0.2277[/tex]
The formula for the binomial distribution is,
[tex]P(X = x) = n_C_x \times px \times (1 - p)^{n-x}[/tex]
Part(a):
Given that,
[tex]n = 20p = 0.27[/tex]
Then from the above formula,
[tex]P(X = 8) = 20_C_8 \times 0.278 \times (1 - 0.27)^{12}\\P(X = 8) = 0.0815[/tex]
Part(b):
Given that,
[tex]n = 20\\p = 0.2[/tex]
Then from the above formula,
[tex]P(X = 0) = {20}_C_0 \times 0.20 \times 0.720 \\P(X = 0) = 0.144[/tex]
Part(c):
Given that,
[tex]n = 20p = 0.3[/tex]
Then from the above formula,
[tex]P(X > 7) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)\\= 1 - (20_C_0 \times 0.30 \times 0.720 + 20_C_1 \times 0.31 \times 0.719 + 20_C_2 \times 0.32 \times 0.718 + 20_C_3 \times 0.33 \times 0.717 + 20_C_4 \times 0.34 \times 0.716 + 20_C_5 \times 0.35 \times 0.715 + 20_C_6 \times 0.36 \times 0.714 + 20_C_7 \times 0.37 \times 0.713\\= 1 - 0.7723\\= 0.2277[/tex]
Learn more about the topic Binomial distribution:
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