1. **Stating the problem:** We want to test if the given frequency distribution data follows a normal distribution using the Chi-Square goodness-of-fit test at significance level $\alpha = 0.05$. 2. **Given data:** - Sample mean $\bar{x} = 109.90$ - Sample standard deviation $s = 19.05$ - Total frequency $n = 40$ - Class intervals and observed frequencies ($f_i$): - 66–77: 2 - 78–89: 4 - 90–101: 7 - 102–113: 10 - 114–125: 8 - 126–137: 6 - 138–149: 3 3. **Formula and rules:** The Chi-Square test statistic is: $$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$$ where $O_i$ is the observed frequency and $E_i$ is the expected frequency under the normal distribution. Expected frequencies $E_i$ are calculated by: $$E_i = n \times P_i$$ where $P_i$ is the probability that a normal variable with mean $\bar{x}$ and standard deviation $s$ falls in the $i$-th class interval. 4. **Calculate $P_i$ for each class interval:** We standardize the class boundaries: $$Z = \frac{X - \bar{x}}{s}$$ Calculate $Z$ for each boundary: - 66: $Z = \frac{66 - 109.90}{19.05} = -2.29$ - 77: $Z = \frac{77 - 109.90}{19.05} = -1.74$ - 89: $Z = \frac{89 - 109.90}{19.05} = -1.11$ - 101: $Z = \frac{101 - 109.90}{19.05} = -0.47$ - 113: $Z = \frac{113 - 109.90}{19.05} = 0.16$ - 125: $Z = \frac{125 - 109.90}{19.05} = 0.79$ - 137: $Z = \frac{137 - 109.90}{19.05} = 1.42$ - 149: $Z = \frac{149 - 109.90}{19.05} = 2.00$ 5. **Find cumulative probabilities from standard normal table:** - $\Phi(-2.29) = 0.0110$ - $\Phi(-1.74) = 0.0418$ - $\Phi(-1.11) = 0.1335$ - $\Phi(-0.47) = 0.3192$ - $\Phi(0.16) = 0.5636$ - $\Phi(0.79) = 0.7852$ - $\Phi(1.42) = 0.9222$ - $\Phi(2.00) = 0.9772$ 6. **Calculate $P_i$ for each interval:** - 66–77: $P_1 = \Phi(-1.74) - \Phi(-2.29) = 0.0418 - 0.0110 = 0.0308$ - 78–89: $P_2 = \Phi(-1.11) - \Phi(-1.74) = 0.1335 - 0.0418 = 0.0917$ - 90–101: $P_3 = \Phi(-0.47) - \Phi(-1.11) = 0.3192 - 0.1335 = 0.1857$ - 102–113: $P_4 = \Phi(0.16) - \Phi(-0.47) = 0.5636 - 0.3192 = 0.2444$ - 114–125: $P_5 = \Phi(0.79) - \Phi(0.16) = 0.7852 - 0.5636 = 0.2216$ - 126–137: $P_6 = \Phi(1.42) - \Phi(0.79) = 0.9222 - 0.7852 = 0.1370$ - 138–149: $P_7 = \Phi(2.00) - \Phi(1.42) = 0.9772 - 0.9222 = 0.0550$ 7. **Calculate expected frequencies $E_i = n \times P_i$:** - $E_1 = 40 \times 0.0308 = 1.23$ - $E_2 = 40 \times 0.0917 = 3.67$ - $E_3 = 40 \times 0.1857 = 7.43$ - $E_4 = 40 \times 0.2444 = 9.78$ - $E_5 = 40 \times 0.2216 = 8.86$ - $E_6 = 40 \times 0.1370 = 5.48$ - $E_7 = 40 \times 0.0550 = 2.20$ 8. **Calculate Chi-Square statistic:** $$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} = \frac{(2-1.23)^2}{1.23} + \frac{(4-3.67)^2}{3.67} + \frac{(7-7.43)^2}{7.43} + \frac{(10-9.78)^2}{9.78} + \frac{(8-8.86)^2}{8.86} + \frac{(6-5.48)^2}{5.48} + \frac{(3-2.20)^2}{2.20}$$ Calculate each term: - $\frac{(0.77)^2}{1.23} = 0.48$ - $\frac{(0.33)^2}{3.67} = 0.03$ - $\frac{(-0.43)^2}{7.43} = 0.02$ - $\frac{(0.22)^2}{9.78} = 0.005$ - $\frac{(-0.86)^2}{8.86} = 0.08$ - $\frac{(0.52)^2}{5.48} = 0.05$ - $\frac{(0.80)^2}{2.20} = 0.29$ Sum: $$\chi^2 = 0.48 + 0.03 + 0.02 + 0.005 + 0.08 + 0.05 + 0.29 = 0.955$$ 9. **Degrees of freedom:** $$df = k - p - 1 = 7 - 2 - 1 = 4$$ where $k=7$ classes, $p=2$ parameters estimated (mean and std dev). 10. **Critical value:** From Chi-Square table for $df=4$ and $\alpha=0.05$, critical value $\chi^2_{0.05,4} = 9.488$. 11. **Decision:** Since calculated $\chi^2 = 0.955 < 9.488$, we fail to reject the null hypothesis. **Conclusion:** The data does not significantly differ from a normal distribution at the 5% significance level. Therefore, the data can be considered normally distributed.