Question

Lie Detector: Suppose a lie detector allows 20% of all lies to go undetected.

(a) If you take the test and tell 10 lies, what is the probability of having 5 or more go undetected? Round your answer to 3 decimal places.

(b) Would 5 undetected lies be an unusually high number* of undetected lies? Use the criteria that a number (x) is unusually large if P(x or more) ≤ 0.05.
-Yes, that is an unusually high number of undetected lies.
-No, that is not an unusually high number of undetected lies.

...............................................

*Unusually low and high number of successes: x successes among n trials is an unusually high number of successes if P(x or more) ≤ 0.05. x successes among n trials is an unusually low number of successes if P(x or fewer) ≤ 0.05.

Answer 1

Given that  a lie detector allows 20% of all lies to go undetected

so p = 0.2

Here the number of undetected lies (X) follows a binomial distribution since there are only 2 outcomes possible here, either detected or undetected

Question (a)

If you take the test and tell 10 lies, what is the probability of having 5 or more go undetected

We need to find P(X 5)

The number of successes ''x'' in ''n'' trails in binomial distribution = * p^x* (1-p)^p-x

= n! / [ x! (n-x)! ] * p^x * (1-p)^n-x

Here n = 10, x = 5

p = 0.2

P(X 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

= 10! / [ 5! (10-5)! ] * 0.2⁵ * (1-0.2)^10-5  + 10! /10! / [ 6! (10-6)! ] * 0.2⁶ * (1-0.2)^10-6  + [ 7! (10-7)! ] * 0.2⁷ * (1-0.2)^10-7  + 10! / [ 8! (10-8)! ] * 0.2⁸ * (1-0.2)^10-8  + 10! / [ 9! (10-9)! ] * 0.2⁹ * (1-0.2)^10-9  + 10! / [ 10! (10-10)! ] * 0.2¹⁰ * (1-0.2)^10-10  

= 0.026424 + 0.005505 + 0.000786 + 0.000073 + 0.0000041 + 0.0000001

= 0.032793

= 0.033 rounded to 3 decimal places

If you take the test and tell 10 lies, what is the probability of having 5 or more go undetected = 0.033

Question (b)

Here P(X 5) = 0.033 which is less than or equal to 0.05

So Yes, that is is an unusually high number of undetected lies.