= Solution
For independent <Poisson random variables> with known expected counts $\mu_i$, the model variance is $\mu_i$. The <Pearson chi-squared statistic> is therefore
$$
\chi_P^2=\sum_i\frac{(n_i-\mu_i)^2}{\mu_i}
=\frac{121}{20}+\frac{121}{35}+\frac{144}{44}+\frac{49}{36}+\frac{144}{24}+\frac{144}{11}
\simeq\boxed{33.23.}
$$
There are six independently specified bins and no parameters fitted from them, giving approximately six <degrees of freedom>. This is far above its expected value six; the approximate <chi-squared distribution> tail <probability> is $9.5\times10^{-6}$. \b[The discrepancies are much too large for a good Poisson-noise fit.] Replacing model variances by the observed-count estimates gives the alternative statistic $30.92$, with the same qualitative conclusion. There is no fitted normalization to justify subtracting a degree of freedom or rescaling the model to the observed total.
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