Answer: The sample variance is \(\boxed{2.65}\) using denominator \(40\). The expected frequency for \(x=4\) is \(6.61\). The Poisson distribution is a suitable model at the 5% level.

(i) From the table,

\[\sum xf=0(3)+1(5)+2(8)+3(10)+4(5)+5(7)+6(2)=118.\]

There are \(40\) days, so the mean is

\[\bar{x}=\frac{118}{40}=2.95,\]

as required.

Also,

\[\sum x^2f=0^2(3)+1^2(5)+2^2(8)+3^2(10)+4^2(5)+5^2(7)+6^2(2)=454.\]

Using denominator \(40\), the variance is

\[\frac{454}{40}-\left(\frac{118}{40}\right)^2.\]

Equivalently,

\[\frac{454-\frac{118^2}{40}}{40}=2.65.\]

A Poisson distribution has mean equal to variance. Here the mean \(2.95\) and variance \(2.65\) are fairly close, so a Poisson distribution might be a reasonable model.

(ii) For a Poisson distribution with mean \(\lambda=2.95\),

\[P(X=4)=\frac{e^{-2.95}2.95^4}{4!}.\]

Since there are \(40\) days, the expected frequency for \(x=4\) is

\[E_4=40\frac{e^{-2.95}2.95^4}{4!}.\]

Evaluating this gives

\[E_4=6.61\]

to 2 decimal places.

(iii) Test

\[H_0:\text{ a Poisson distribution with mean }2.95\text{ fits the data}.\]

Some expected frequencies are less than 5, so combine classes. A suitable grouping is:

\[\begin{array}{c|ccccc} \text{class} & 0,1 & 2 & 3 & 4 & 5,6,\geq7\\ \hline O & 8 & 8 & 10 & 5 & 9\\ E & 8.27 & 9.11 & 8.96 & 6.61 & 7.05 \end{array}\]

The chi-squared statistic is

\[\chi^2=\sum\frac{(O-E)^2}{E}.\]

So

\[\chi^2=\frac{(8-8.27)^2}{8.27}+\frac{(8-9.11)^2}{9.11}+\frac{(10-8.96)^2}{8.96}+\frac{(5-6.61)^2}{6.61}+\frac{(9-7.05)^2}{7.05}.\]

This gives

\[\chi^2=1.20\quad\text{to 3 significant figures}.\]

There are 5 combined cells. One parameter, \(\lambda\), has been estimated from the data, so the degrees of freedom are

\[5-1-1=3.\]

At the 5% significance level, the critical value is

\[\chi^2_{3,0.95}=7.815.\]

Since

\[1.20<7.815,\]

we do not reject \(H_0\). Therefore the Poisson distribution is a suitable model for the data.