Question

Alex is back in the game on Tinder. Let ?? be a random variable that is equal to 1 if?-th person is a match. Suppose each match happens independently of the others with probability 0.3. Let ?₁, . . . , ?? denote a random sample.

1) If ? = 6, ? (X̄? ≤ 1/6) is equal to

(a) 0.34

(b) 0.36

(c) 0.42 (correct)

(d) 0.44

2) If ? = 6, ? (X̄? ≤ 5/6) is equal to

(a) 0.923

(b) 0.678

(c) 0.111

(d) 0.999 (correct)

3) The variance ?² of ?? is
(a) 0.21 (correct)

(b) 0.3

(c) 0.09

(d) 0.12

4)If ? = 36, ? (X̄? ≤ 0.35) is approximately equal to

(a) 0.6546

(b) 0.7422 (correct)

(c) 0.8216

(d) 0.9347

5) If ? = 36, ? (X̄? ≤ 0.25) is approximately equal to

(a) 0.2578 (correct)

(b) 0.3244

(c) 0.7422

(d) 0.8216

Answer 1

1)for binomial distribution

P(Xbarn <=1/6)=P(X <=6/6)=P(X <=1) =P(X=0)+P(X=1)=⁶C₀(0.3)⁰(0.7)⁶+⁶C₁(0.3)¹(0.7)⁵=0.4202

2) ? (X̄? ≤ 5/6) =P(X<=5)=1-P(X=6)=1-⁶C₆(0.3)⁶(0.7)⁰ =1-0.0007 =0.9993

3) variance =p(1-p)=0.3*(1-0.3)=0.21

4)

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.300
sample size       =n=	36
std error of proportion=σp=√(p*(1-p)/n)=	0.0764

probability =

P(X<0.35)

=

P(Z<0.65)=

0.7422

5)

probability =

P(X<0.25)

=

P(Z<-0.65)=

0.2578