Expected Value = np = 14(0.23) = 3.22
Standard Deviation = √(npq) = √[14(0.23)(0.77)] ≈ 1.5746
Z-score = (x-µ)/σ = (4-3.22)/1.5746 ≈ 0.5
P(Z≤0.50) ≈ 0.6915
Therefore, the probability that 4 students at most will attend the Tet festivities is about 0.6915
Hope this helped!