Question

H al Random Variable n=30 2. It has been reported that 39% of college students graduate in 4 years. Consider a random sample
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Answer #1

Answer:

Given,

sample n = 30

p = 39% = 0.39

Mean = np = 30*0.39 = 11.7

standard deviation = sqrt(npq) = sqrt(30*0.39*0.61) = 2.6715

a)

Consider P(X) = nCr*p^r*q^(n-r)

P(X = 13) = 30C13*0.39^13*(1-0.39)^(30-13)

= 119759850*0.39^13*0.61^17

= 0.1296

b)

P(X <= 15) = P((x-u)/s < (15.5 - 11.7)/2.6715)

= P(z < 1.4224)

= 0.9225449 [since from z table]

= 0.9225

c)

P(X > 15) = 1 - P(X <= 15)

= 1 - 0.9225

= 0.0775

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