Question

9. Discrete probability distributions #1

Who’s so vain? A survey conducted by the American Association of Motor Vehicle Administrators (AAMVA) and Stefan Lonce, author of LCNS2ROM—License to Roam: Vanity License Plates and the Stories They Tell, reveals that Virginia motor vehicle owners are the vainest. Approximately 16% of Virginia license plates are vanity plates.

Select the appropriate distribution in the Distributions tool to help answer the questions that follow.

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You randomly select 25 Virginia license plates. Let X denote the number of vanity plates.

The probability that exactly four license plates are vanity plates is--------------- .

The probability that at least seven license plates are vanity plates is------------- .

The expected value of X is 4.0   , and the standard deviation of X is 1.8330   

Answer 1

Binomial distribution

a)

Here, n = 25, p = 0.16, (1 - p) = 0.84 and x = 4
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 4)
P(X = 4) = 25C4 * 0.16^4 * 0.84^21
P(X = 4) = 0.213
0

b)

Here, n = 25, p = 0.16, (1 - p) = 0.84 and x = 7
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 6).
P(X <= 6) = (25C0 * 0.16^0 * 0.84^25) + (25C1 * 0.16^1 * 0.84^24) + (25C2 * 0.16^2 * 0.84^23) + (25C3 * 0.16^3 * 0.84^22) + (25C4 * 0.16^4 * 0.84^21) + (25C5 * 0.16^5 * 0.84^20) + (25C6 * 0.16^6 * 0.84^19)
P(X <= 6) = 0.0128 + 0.0609 + 0.1392 + 0.2033 + 0.213 + 0.1704 + 0.1082
P(X <= 6) = 0.9078

P(X> =7) = 1 - P(x< =6)
= 1 - 0.9078
= 0.0922

c)

Expected value = np
= 25 *0.16 =4

std.deviation = sqrt(npq)
= sqrt(25 * 0.16 *(1-0.16))
= 1.8330