Answer:
The answer is below
Step-by-step explanation:
given that:
n = sample size = 200, p = probability of a cancelled reservation = 1 - 0.9 = 0.1
The z score is given by:
[tex]z=\frac{x-mean}{standard\ deviation} =\frac{x-np}{\sqrt{np(1-p)} }[/tex]
a) The probability that all the passengers who arrives can be seated can only occur if atleast 15 or more passengers cancelled their reservations.
P(X ≥ 15) = 1 - P(X ≤ 14.5)
[tex]z=\frac{x-np}{\sqrt{np(1-p)} }=\frac{14.5-(200*0.1)}{\sqrt{(200*0.1)(1-0.1)} }=-1.3[/tex]
From the normal distribution table:
P(X ≥ 15) = 1 - P(X ≤ 14.5) = 1 - P(z < -1.3) = 1 - 0.0968 = 0.9
b) The probability that there are empty seat only occur if atleast 16 or more passengers cancelled their reservations.
P(X ≥ 16) = 1 - P(X ≤ 15.5)
[tex]z=\frac{x-np}{\sqrt{np(1-p)} }=\frac{15.5-(200*0.1)}{\sqrt{(200*0.1)(1-0.1)} }=-1.06[/tex]
From the normal distribution table:
P(X ≥ 16) = 1 - P(X ≤ 15.5) = 1 - P(z < -1.06) = 1 - 0.1446 = 0.8554
c)
[tex]0.95=P(X\geq n-185)=1-P(X\leq n-185)\\\\P(X\leq n-185)=1-0.95\\\\P(X\leq n-185)=0.05\\\\0.05=P(Z<\frac{n-185-0.1n}{\sqrt{0.9*0.1*n} } )\\\\0.05 \ correspond\ to \ a\ probability\ of\ -1.64\\\\-1.64=\frac{0.9n-185}{\sqrt{0.9*0.1*n} } \\\\n=198[/tex]
