Question 1125237
The SE is sd/sqrt(35)=5.14
Now the z-value (if this is a known sd, which it sounds like from the wording)is (140-150)/5.14 to (191-150)/5.14
z=(x bar-mean)/sigma/sqrt(n)
-1.95 to 7.98
probability is 0.9744

This says in essence that the likelihood of the MEAN of a sample being in this region is much higher than a single value.