Question

(1 point) Use normal approximation to estimate the probability of getting at most 45 girls in 100 births. Assume that boys and girls are equally likely. (1 point) A coin is tossed 12 times. b) What is the probability of getting exactly 2 heads? (1 point) An airline company is considering a new policy of booking as many as 336 persons on an airplane that can seat only 300. (Past studies have revealed that only 81% of the booked passengers actually arrive for the flight.) Estimate the probability that if the company books 336 persons, not enough seats will be available.

Answer 1

1. PROBABILITY OF BOY BIRTH =PROBABILITY OF GIRL BIRTH=$\frac{1}{2}=0.5$

X=a girl is born

p=0.5(success)

q=1-p=1-0.5=0.5

n=100

np=100*0.5=50

npq=100*0.5*0.5=25

using normal approximation:

X~N(np,npq)

X~N(50,25)

P(X$\leq$45)=P(Z<$\frac{45-50}{\sqrt{25}}$)

=P(Z<-1)

P(X$\leq$45)=0.1587

(b)

n=12

probability of head=probability of tail=$\frac{1}{2}=0.5$

X=getting heads

p=0.5(success)

q=1-p=1-0.5=0.5

X~Binom(12,0.5)

P(X=2)=$^{12}C_{2}*(0.5)^2*(0.5)^{10}$

P(X=2)=66*0.25*0.0009765

P(X=2)=0.01611

(c)

n=336

p(success)=81%=0.81

q(failure)=1-p=1-0.81=0.19

mean=np=336*0.81=272.16

variance=npq=336*0.81*0.19=51.71

standard deviation=$\sqrt{51.71}$=7.19

X~Binom(336,0.81)

mean=np=336*0.81=272.16

variance=npq=336*0.81*0.19=51.71

standard deviation=$\sqrt{51.71}$=7.19

P(not enought seats will be available)=P(300$\leq$ X $\leq$ 336)

=P(X$\leq$336)-P(X$\leq$300)

=P(Z$\leq$$\frac{336-272.16}{7.19}$)-P(Z$\leq$$\frac{300-272.16}{7.19}$)

=P(Z$\leq$8.87)-P(Z$\leq$3.87)

=1-0.99998725

P(300$\leq$ X $\leq$ 336)=0.00001275

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