Answer:
P(Exactly 1 is being deceptive) is [tex]\simeq[/tex] 0.2062 .
P(At most 1 is being deceptive) is [tex]\simeq 0.2749[/tex]
Mean of the distribution is, 2.4 and standard deviation of the distribution is,
[tex]\simeq 0.4382[/tex]
Step-by-step explanation:
Let, the no. of truthful persons suggested as deceptive by the lie-detector test be denoted by the random variable X. Then, according to the question, in this case,
X [tex]\sim[/tex] Binomial (12, 0.2)
So, here,
1. No. of trials = 12 = n (say)
2. Probability of success = 0.2 = p (say)
3. Probability of failure = (1 - 0.2) = 0.8 = q (say)
So,
P(Exactly 1 is being deceptive)
= P(X = 1)
= [tex]^{12}C_{1} \times (0.2)^{1} \times (0.8)^{11}[/tex]
[tex]\simeq[/tex] 0.2062 ---------------(1)
P(At most 1 is being deceptive)
= P(X = 0) + P(X = 1)
=[tex]\sum_{x = 0}^{1}(^{12}C_{x}\times (0.2)^{x} \times (0.8)^{(12 - x)}[/tex]
[tex]\simeq ^{12}C_{0} \times (0.2)^{0} \times (0.8)^{12} + 0.2062[/tex]
[From (1) putting the value of P(X = 1)]
[tex]\simeq (0.0687 + 0.2062)[/tex]
= 0.2749
Mean of the distribution = [tex]n \times p[/tex]
= [tex]12 \times 0.2[/tex]
= 2.4
Standard deviation of the distribution,
=[tex]\sqrt {n \times \p \times q}[/tex]
=[tex]\sqrt {12 \times 0.2 \times 0.8}[/tex]
[tex]\simeq 0.4382[/tex]
