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2 answers:
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Answer: 0.8461
Step-by-step explanation:
Given :
Let x be the random variable that represents the cost for the hospital emergency room visit.
We assume that cost for the hospital emergency room visit is normally distributed .
z-score for x=1000 ,
Using z-value table , we have
P-value =P(x>1000)=P(z>-1.02)=1-P(z≤ -1.02)=1-0.1538642
=0.8461358≈0.8461 [Rounded nearest 4 decimal places]
Hence, the probability that the cost will be more than $1000 = 0.8461