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Lv 6
? asked in Education & ReferenceHomework Help · 8 years ago

Statistics final review- please help me learn how to do this problem?

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. Thanks!!!

Update:

@Bee- could you slow it down in the beginning? i dont see how you got 4.62 thanks1

Update 2:

@Bee- could you slow it down in the beginning? i dont see how you got 4.62 thanks1

Update 3:

@Bee- could you slow it down in the beginning? i dont see how you got 4.62 thanks1

Update 4:

@Bee- could you slow it down in the beginning? i dont see how you got 4.62 thanks1

2 Answers

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  • 8 years ago
    Favorite Answer

    negative

    Here is a link to the table of "continuity corrections" . You need to "memorize" these for your test ...

    http://people.richland.edu/james/lecture/m170/ch07...

    p = 0.76

    mean = np = 117(0.76) = 88.92

    s.d. = sqrt(npq) = sqrt[(117)(0.76)(1-0.76)] = 4.62

    Now use the Normal Approximation with continuity corrections ...

    A) P(X = 98) = P(97.5 < X < 98.5) ... now convert to z-values:

    P[(97.5 - 88.92) / (4.62) < z < (98.5 - 88.92) / (4.62)]

    P( 1.86 < z < 2.07) = .0122

    Use the same method and the table of "continuity corrections" to solve B, C and D.

    Hope that helps

  • ?
    Lv 4
    5 years ago

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