Question
one question 2 pictures
2.4.8, pp 97 101. A student counts the number of chocolate chips in cookies and records the data shown in the table below (da
none of these x 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 freq 1 1 1 2 2 0 0 5 5 1 0 1 1 0 0 rel freq 0.04 0.05 0.06 0.07
0 0
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Answer #1

Fitting Poisson Distribution

We get the relative frequencies by dividing each frequency by the total frequency. If the number of choco chips is described by a Poisson distribution, then \lambda should be equal to the mean of the sample. We get the following table:

x frequency(f) relative freq xf
16 1 0.05 16
17 1 0.05 17
18 1 0.05 18
19 2 0.1 38
20 2 0.1 40
21 0 0 0
22 0 0 0
23 5 0.25 115
24 5 0.25 120
25 1 0.05 25
26 0 0 0
27 1 0.05 27
28 1 0.05 28
29 0 0 0
30 0 0 0
Total 20 1 444

So the mean is given by: ΣΤΣΤ = 444/20 = 22.2.

Thus \lambda\approx 22.2.

Now we will calculate the mass values.

We know that the pmf of Poisson distribution is given by: f(x) = exp(-\lambda).\lambdax/ x!

So putting \lambda\approx 22.2 we get:

x f(x)
16 0.04
17 0.05
18 0.06
19 0.07
20 0.08
21 0.08
22 0.09
23 0.08
24 0.08
25 0.07
26 0.06
27 0.05
28 0.04
29 0.03
30 0.02

Thus third option is the correct one.

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