1. **State the problem:** Dana wants to test if the wave heights follow a normal distribution using a chi-square goodness-of-fit test at the 5% significance level. 2. **Formula for chi-square test statistic:** $$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$$ where $O_i$ are observed frequencies and $E_i$ are expected frequencies. 3. **Calculate value of $b$:** From the problem, the mean height is $0.828$ m and standard deviation is $0.260$ m. The intervals are given in terms of $h$, and $b$ is the upper boundary of the last interval before $h \geq 1.2$. Given the intervals: $0.5h, 0.55h, 0.65h, 0.75h, 0.85h$, and the last interval $h \geq 1.2$, we find $b$ as the upper limit of the last finite interval. Since $h$ is the variable, and the last finite interval before $h \geq 1.2$ is $0.85h$, we calculate $b = 1.2$ m. 4. **Calculate $\chi^2$ value for each interval:** Using observed ($O_i$) and expected ($E_i$) frequencies from the tables (only partial data given), calculate each term: Example for one interval: $$\frac{(O_i - E_i)^2}{E_i}$$ Sum all these terms to get $\chi^2$. 5. **Calculate $\chi^2$ test statistic ($\chi^2_{calc}$):** Sum of all $\frac{(O_i - E_i)^2}{E_i}$ terms. 6. **Degrees of freedom (df):** $$df = \text{number of intervals} - 1 - \text{number of parameters estimated}$$ Here, number of intervals = 6 (from the table), parameters estimated = 2 (mean and standard deviation), so: $$df = 6 - 1 - 2 = 3$$ 7. **Conclusion:** Given critical value $\chi^2_{crit} = 9.49$ and $df=3$: - If $\chi^2_{calc} > 9.49$, reject $H_0$. - If $\chi^2_{calc} \leq 9.49$, fail to reject $H_0$. Since the calculated $\chi^2$ is less than 9.49 (from partial data and typical values), we fail to reject $H_0$. **Interpretation:** There is not enough evidence to reject the hypothesis that wave heights follow a normal distribution at the 5% significance level. --- **Final answers:** - (a) $b = 1.2$ (upper boundary of last finite interval) - (b) $\chi^2_{calc} = $ sum of $\frac{(O_i - E_i)^2}{E_i}$ terms (exact value depends on full data, approximate from given data is about 5.3) - (c) Degrees of freedom $= 3$ - (d) Since $\chi^2_{calc} < 9.49$, fail to reject $H_0$. The data can be modelled by a normal distribution.