Question

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 hasa Poisson distribution with parameter µ. Use the accompanying data on absences for 50 days to obtain a 95% large-sample CI for µ.
Number of
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 µ, so
Z = (X - µ)/square root(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) andsolving the resulting inequalities for µ.] (Round your answers to two decimal places.)

Answer 1

From the given data we have

x fi fixi
0 2 0
1 3 3
2 8 16
3 11 33
4 8 32
5 7 35
6 6 36
7 2 14
8 1 8
9 1 9
10 1 10
n= 50 196

Mean λ=196/50

Answer 2

Take the frequencies and number of absences and tally them. They will tally to 196. Then divide 196 by 50 -- 3.92. This is x(bar). Then use the equation Z=(X(bar) - mu) / sqrt(mu/n) and insert the values for x(bar) and n. Multiply both sides by the square-root - 1.96=(3.92 - mu)/sqrt(mu/50) => 1.96 * sqrt(mu/50) = (3.92 - mu). Square both sides => (1.96 * sqrt(mu/50))^2 = (3.92 - mu)^2, to clear the square-root and solve for mu. 

The solve function like the one in the TI-89 calculator, replacing x for mu, will do this for you:

solve((1.96 * sqrt(x/50)^2 = (3.92 - x)^2, x)

mu = 3.40827 or mu = 4.05856 (re-inserting mu for x.)