A certain flight arrives on time
83 percent of the time. Suppose 130 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that
(a) exactly 117 flights are on time.
(b) at least 117
flights are on time.
(c) fewer than 96 flights are on time.
(d) between 96 and 108, inclusive are on time.
Mean = np = 130(0.83) = 107.9
Standard deviation = = 4.283
Hence,
a) P(X = 117)
= P(116.5 < X < 117.5) [Continuity Correction]
= P(2.01 < z < 2.24)
= 0.0097
b) P(Atleast 117)
= P(X > 116.5) [Continuity Correction]
= P(z > 2.01)
= 0.0222
c) P(Fewer than 96)
= P(X < 95.5) [Continuity Correction]
= P(z < -2.90)
= 0.0019
d) P(Between 96 and 108, inclusive)
= P(95.5 < X < 108.5) [Continuity Correction]
= P(-2.90 < z < 0.14)
= 0.5538
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