Answer
$0.113$
between 4869 and 5679
Work Step by Step
Given $\mu=5274, \sigma=600$
Value 6000 corresponds to $z=\frac{6000-5274}{600}=1.21$
The area to the left of z=1.21 is 0.887
The probability that a randomly selected person spends more than 6000 is
$1-0.887=0.113$
Middle $50\%$ corresponds to z scores from -0.6745 to +0.6745,
this gives the range of expenditures from $5274-600\times0.6745$ to $5274+600\times0.6745$
So the limits are 4869 and 5679
