Question 1178240
the short answer that is usually given is yes, because the sample size is > 30. One has to remember, however, that with a normal distribution, a sample size of 10 is sufficient and with a skewed population, a sample size of 50 may not be.

Anyway, given that, the sd of the sample means (the standard error) is sigma/sqrt(42)=1.389
The probability between 53 and 54 uses z=(53-52)/9/sqrt(42) or 0.72 and z=2/9/sqrt(42)=1.44
The probability of z being between 0.72 and 1.44 is 0.1608.

z(0.45)=-0.1257
z=(x bar-mean)/sd/sqrt(n)
-0.1257=(x bar-52/1.389
-0.1745=xbar-52
xbar=51.8255 or 51.8