Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
statistics review for final?
A certain flight arrives on time 76 percent of the timre. Suppose 117 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that
A exactly 98 flights are on time
B at least 98 flights are on time
C fewer than 75 flights are on time
D between 75 and 101 flights are on time. The answers are .0122, .0314, .0009, .9958. I just need to know how they got them.
1 Answer
- Julius NLv 78 years agoFavorite Answer
I get slightly different answers, maybe their tables are not as precise as my calculator?
n = 117; p = 0.76; q = 0.24; P(x ) = nCx p^x (1-p)^(n-x)
μ = (n p) = 88.92; σ² = ( n p q ) = 21.3408; σ = √( n p q ) = 4.6196
A:
binomial answer: P(98 ) = nCx p^x (1-p)^(n-x) = 0.0121078
normal approximation: z = ( x - μ ) / σ
P (97.5 < x < 98.5) = P (1.8573 < z < 2.0738) = 0.0126
B. P (x > 97.5) = P (z > 1.8573) = 0.0316
C. P ( x < 74.5) = P ( z < -3.1215 ) = 0.0009
D. P (74.5 < x < 101.5) = P (-3.1215 < z < 2.7232) = 0.9959