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24 March, 19:38

The superintendent of a large school district, having once had a course in probability and statistics, believes that the number of teachers absent on any given day has a Poisson distribution with parameter µ.

Use the accompanying data on absences for 50 days to obtain a 95% large sample CI for µ.

Absences: 0 1 2 3 4 5 6 7 8 9 10

Frequency: 2 3 8 11 8 7 6 2 1 1 1

[Hint: The mean and variance of a Poisson variable both equal m, so

Z = (X - µ) / √ (µ/n),

has approximately a standard normal distribution. Now proceed as in the derivation of the interval for p by making a probability statement (with probability 1 - a) and solving the resulting inequalities for µ.

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  1. 24 March, 19:52
    0
    The confidence interval required is (3.37,4.47)

    Step-by-step explanation:

    From the given data

    Let X represents the absence and f is the frequency

    Summation of f = 50

    Summation of fX=196

    Mean of a Poisson distribution = 196/50 = 3.92

    Critical value

    At 5% level of significance for a two tailed z-distribution

    Z (0.05/2) = + or - 1.96

    The 95% confidence interval sample confidence interval

    Cl (95%) = mean+or - z (at 0.05/2) * √ (mean/n

    Where n=50

    Cl (95%) = 3.92+/-1.96 (√3.92/50)

    = (3.37,4.47)
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