Question

A sociology class contains 10 Bernie fans and 8 Hilary fans. What is the probability of choosing 5 at random and having 2 Hilary fans?

B.How big of sample is need to estimate the mean time with a 95% confidence interval and a maximum error of 1? The time has a variance equal to 20.

C, The probability that a student likes statistics is .1. What is the probability that 5 or less students like stats in a class of 16?

Answer 1

(A) P(Event) = Number of favourable outcomes/Total Number of outcomes

Please note ⁿC_x = n! / [(n-x)!*x!]

Total fans = 10 + 8 = 18. Therefore Total outcomes = choosing 5 fans out of 15 = ¹⁸C₅ = 8568

Favourable outcomes: Choosing 2 Hilary fans out of 8 and choosing 3 bernie fans out of 10

= ⁸C₂ * ¹⁰C₃ = 28 * 120 = 3360

Therefore the required probability = 3360/8568 = 20/51 = 0.3922

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(B)

Given ME = 1, = 20, = 0.05

The Zcritical at = 0.05 is 1.96

The ME is given by :

ME = Zcritical * \frac{\sigma}{\sqrt{n}}

Squaring both sides we get: (ME)² = (Z critical)² * /n

Therefore n = (Zcritical /ME)² * = (1.96*/1)² * 20 = 76.832

Therefore n = 77 (Taking it to the next whole number)

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Binomial Probability = ⁿC_x * (p)^x * (q)^n-x, where n = number of trials and x is the number of successes.

Here n = 16, p = 0.1, q = 1 – p = 0.9.

To Find P( X 5) = P(5) + P(4) + P(3) + P(2) + P(1) + P(0)

P(X = 5) = ¹⁶C₅ * (0.1)⁵ * (0.9)^{16-5 = 11} = 0.0137

P(X = 4) = ¹⁶C₄ * (0.1)⁴ * (0.9)^{16-4 = 12} = 0.0514

P(X = 3) = ¹⁶C₃ * (0.1)³ * (0.9)^{16-3 = 13} = 0.1423

P(X = 2) = ¹⁶C₂ * (0.1)² * (0.9)^{16-2 = 14} = 0.2745

P(X = 1) = ¹⁶C₁ * (0.1)¹ * (0.9)^{16-1 = 15} = 0.3294

P(X = 0) = ¹⁶C₀ * (0.1)⁰ * (0.9)^{16-0 = 16} = 0.1853

Therefore P( X 5) = 0.0137 + 0.0514 + 0.1423 + 0.2745 + 0.3294 + 0.1853 = 0.9966

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