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Answer: 0.0174
Step-by-step explanation:
Given :
Let x be the random variable that represents the weight of the drumsticks.
We assume that the weight of the drumsticks is normally distributed.
Now, the z-score for x=2.23 ,
Using z-value table , we have
P-value =P(x≥2.23)=P(z≥2.11)=1-P(z<2.11)=1-0.9825708
=0.0174292≈0.0174 [Rounded nearest 4 decimal places]
Hence, the probability of the stick's weight being 2.23 oz or greater = 0.0174