Can someone help walk me through solving this problem? I keep getting it wrong

Question

1. In 1940 the average size of a U.S. farm was 174 acres. Let's say that the standard deviation was 51 acres. Suppose we randomly survey 45 farmers from 1940

I need help solving the question below.

"The middle 50% of the distribution for X, the bounds of which form the distance represented by the IQR, lies between what two values? (answer must be rounded to two decimal places)"

Answer 1

find 25th and 75th percentiles

assume data normally distributed N(174,51)

z = (x - mean)/standard deviation is N(0,1)

z = -0.67 for 25th percentile

find x for 25th percentile

-0.67 = (x - 174)/51

z = 0.67 for 25th percentile

find x for 75th percentile

0.67 = (x - 174)/51