Determine whether the normal distribution can be used to compare the following population proportions. \[ \mathrm{n}_{1}=26, \quad \mathrm{n}_{2}=38, \quad \hat{\mathrm{p}_{1}}=0.923, \quad \hat{\mathrm{p}_{2}}=0.789 \] Step 1 of 2 : Calculate the four values $\mathrm{n}_{1} \hat{\mathrm{p}_{1}}, \mathrm{n}_{1}\left(1-\mathrm{p}_{1}\right), \mathrm{n}_{2} \hat{\mathrm{p}}_{2}$, and $\mathrm{n}_{2}\left(1-\hat{p}_{2}\right)$. Round your answers to three decimal places, if necessary. Answer 2 Points

Step 1 ：Calculate the four values: \(n_{1} \hat{p}_{1}\), \(n_{1}(1-\hat{p}_{1})\), \(n_{2} \hat{p}_{2}\), and \(n_{2}(1-\hat{p}_{2})\)

Step 2 ：For \(n_{1} \hat{p}_{1}\), calculate \(26 * 0.923\) to get \(23.998\)

Step 3 ：For \(n_{1}(1-\hat{p}_{1})\), calculate \(26 * (1-0.923)\) to get \(1.992\)

Step 4 ：For \(n_{2} \hat{p}_{2}\), calculate \(38 * 0.789\) to get \(29.982\)

Step 5 ：For \(n_{2}(1-\hat{p}_{2})\), calculate \(38 * (1-0.789)\) to get \(8.018\)

Step 6 ：Round these four values to three decimal places: \(n_{1} \hat{p}_{1} = 24.000\), \(n_{1}(1-\hat{p}_{1}) = 2.000\), \(n_{2} \hat{p}_{2} = 30.000\), and \(n_{2}(1-\hat{p}_{2}) = 8.000\)

Step 7 ：Since all four values are greater than or equal to 5, the normal distribution can be used to compare the population proportions. \(\boxed{\text{Yes}}\)