Question

In 2005, 0.76% of all airline flights were on-time. If we choose a simple random sample of 2000 flights, find the probability that... (to four decimal places, using normal chart, no continuity correction)
(a) at least 79% of the sample's flights were on time

(b) at most 1580 of the sample's flights were on time

(c) the sample proportion of on-time flights (p-hat) differs from the truth by more than three percent

Answer 1

Answer:

Step-by-step explanation:

Here X no of flight that reach on time has two outcomes only and each trial is independent of the other.

Since sample size is large we approximate to normal with mean =np and variance =npq

i.e. X is N(152,19.10)

$Z=\frac{x-152}{19.1}$ is used for converting to std normal variate

Continuity correction of 0.5 on either side of interval would be applied for x

a) $P(X\geq 79%) = P(X\geq 1580)\\=P(X\geq 1579.5)\\=P(Z\geq 3.12)$

=0.00

b)$P(X\leq 1580)\\=P(X\leq 1580.5)\\=P(Z\leq 3.17)\\=$=1.00

c) p is N(0.76, sqrtpq/n)=N(0.76,0.0095)

P(p diff>0.03) =

$P(Z>\frac{0.03}{0.0095} \\=P(Z>3.16)$=0.00