Answer: 0.3821
Step-by-step explanation:
Let x be the random variable that represents the wages.
Given: Workers have an average wage of $9.00 per hour with a standard deviation of $0.50.
The probability of obtaining a sample mean less than or equal to $8.85 per hour:
[tex]P(X<8.85)=P(\dfrac{X-mean}{standard\ deviation}<\dfrac{8.85-9.00}{0.50})\\\\=P(Z<-0.3)\\\\=1-P(0.3)\\\\=1-0.6179\\\\=0.3821[/tex]
The required probability = 0.3821