An airline experiences a no-show rate of 6%. What is the maximum number of reservations that it could accept for a flight with a capacity of 160, if it wants the probability of accommodating all reservation holders to be at least

Step-by-step explanation:

Use the normal approximation to the binomial distribution

mean µ = np

standard deviation σ = √npq

Where,

n is sample size

p is probability of success.

q is probability of failure

Given that

q = 6% =0.06

Then, p = 1-q = 1-0.06 = 0.94

Therefore:

µ = pn

µ = 0.94n

Also

σ = √npq

σ = √(n)(0.94)(0.06)

σ = √(.0564n)

Using z-scores:

z = (x — µ )/σ

Using the data above

1.645 = (160 — 0.94n)/√(0.0564n)

Cross multiply

1.645√0.0564n = 160—0.94n

Square both sides

1.645²× 0.0564n = (160-0.94n)

0.153n=25600— 300.8n + 0.8836n²

0.8836n²-300.8n-0.153n +25600=0

0.8836n² — 300.953n + 25600 = 0

Using quadratic formula method.

a = 0.8836 b = -300.953 c = 25600

n = [-b±√(b²-4ac)]/2a

n = [--300.953±√((-300.953)²-4×0.8836×25600)] / (2 × 0.8836)

n = [300.953±√(92.07)]/1.7672

n = (300.953±9.6)/1.762

n = (300.953-9.6)/1.762

n = 168.22

Or

n = (300953+9.6)/1.762

n = 176.25

The maximum number of reservation is approximately 168.

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step-by-step explanation:

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y=3n + 1

step-by-step explanation:

this is wrong because it says the second number is one less and i added one instead