Using the normal distribution principle ; the probability that a randomly chosen chocolate is in the celebration box is 0.553
Zscore = (X - μ) ÷ σ
For X = 5 :
Zscore = (5 - 5.8) ÷ 1.3 = - 0.615
For X = 7 :
Zscore = (7 - 5.8) ÷ 1.3 = 0.923
Using a normal distribution table :
P(Z < 0.923) = 0.822
P(Z < - 0.615) = 0.269
P(Z < 0.923) - P(Z < - 0.615) = 0.822 - 0.269 = 0.553
Therefore, the probability of inclusion is 0.553.
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