Answer: 0.8461
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 ,
![z=(1000-1400)/(392)\approx-1.02\ \ \ [\because z=(x-\mu)/(\sigma)]](https://img.qammunity.org/2020/formulas/mathematics/college/yzpa0qojfizkc7y7p3jjw7bf9jfhwgcczo.png) 
 
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