Solution :
P(winning Ghc 1) = P(getting at least 1 heads in two tosses)= 3 4 =\frac34 = 4 3
Now, n = 300 , p = 3 4 , q = 1 4 n=300,p=\frac34,q=\frac14 n = 300 , p = 4 3 , q = 4 1
μ = n p = 225 , σ = n p q = 56.25 = 7.5 \mu=np=225,\sigma=\sqrt{npq}=\sqrt{56.25}=7.5 μ = n p = 225 , σ = n pq = 56.25 = 7.5
X ∼ N ( μ , σ ) X\sim N(\mu,\sigma) X ∼ N ( μ , σ )
(a)P ( X ≥ 250 ) = 1 − P ( X < 250 ) = 1 − P ( z < 250 − 225 7.5 ) P(X\ge250)=1-P(X<250)=1-P(z<\frac{250-225}{7.5}) P ( X ≥ 250 ) = 1 − P ( X < 250 ) = 1 − P ( z < 7.5 250 − 225 )
= 1 − P ( z < 3.3 ) = 1 − 0.99957 = 0.00043 =1-P(z<3.3)=1-0.99957=0.00043 = 1 − P ( z < 3.3 ) = 1 − 0.99957 = 0.00043
(b) P ( X ≥ 250 ) ≥ 0.99 P(X\ge250)\ge0.99 P ( X ≥ 250 ) ≥ 0.99
⇒ 1 − P ( X < 250 ) ≥ 0.99 ⇒ 0.01 ≥ P ( X < 250 ) ⇒ P ( z < 250 − n ( 3 / 4 ) n ( 3 / 4 ) ( 1 / 4 ) ) ≤ 0.01 \Rightarrow 1-P(X<250)\ge0.99
\\ \Rightarrow 0.01\ge P(X<250)
\\ \Rightarrow P(z<\frac{250-n(3/4)}{\sqrt{n(3/4)(1/4)}})\le 0.01 ⇒ 1 − P ( X < 250 ) ≥ 0.99 ⇒ 0.01 ≥ P ( X < 250 ) ⇒ P ( z < n ( 3/4 ) ( 1/4 ) 250 − n ( 3/4 ) ) ≤ 0.01
⇒ 250 − n ( 3 / 4 ) n ( 3 / 4 ) ( 1 / 4 ) > 0.50399 ⇒ 250 − n ( 3 / 4 ) 0.50399 > ( 3 n / 16 ) ⇒ ( 250 − n ( 3 / 4 ) 0.50399 ) 2 > 3 n / 16 \Rightarrow \frac{250-n(3/4)}{\sqrt{n(3/4)(1/4)}}>0.50399
\\ \Rightarrow \frac{250-n(3/4)}{0.50399}>\sqrt{(3n/16)}
\\ \Rightarrow (\frac{250-n(3/4)}{0.50399})^2>3n/16 ⇒ n ( 3/4 ) ( 1/4 ) 250 − n ( 3/4 ) > 0.50399 ⇒ 0.50399 250 − n ( 3/4 ) > ( 3 n /16 ) ⇒ ( 0.50399 250 − n ( 3/4 ) ) 2 > 3 n /16
On solving, we get,
n > 338.68 n>338.68 n > 338.68
Or n ≥ 339 n\ge339 n ≥ 339
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