What is the probability that the length of the yellowtail is more than 41 inches? Original Question: 


*

*In a survey conducted by the Big Creek State Marine Reserve in California,
the length of various species of fish caught at two locations were recorded.
Suppose the length of a yellowtail caught at the Santa Lucia Kelp Bed is normally distributed with mean 38 inches and standard deviation 2.9 inches.


*

*Suppose 1 yellowtail is selected at random. What is the probability that
the length of the yellowtail is more than 41 inches?

*Suppose 32 yellowtail are selected at random. What is the probability
that the sample mean length is greater than 41 inches?



Now from the question I can see that We are given: Mean = 38 and SD = 2.9. 
From that This is All I can think of --> Z = (41-38)/2.9 = 1.03 = 1 - 0.84849 = 0.15151
Now IDK if 0.15151 is correct or not. So Can someone please help me figure out this question.
Thank you. 
 A: a) You're looking to find the probability that one yellowtail's length is greater than 41.  Since the data follows a normal distribution, you can determine this value via the normalcdf() function on your calculator or on a website.  You enter the mu (38), standard deviation (2.9), and 'above 41.'  It draws a normal distribution and finds the area under the distribution between 41 and positive infinity.
There are plenty of tutorials online on how to do this on your TI-84, or whatever calculator you use.  Just google how to do the normalcdf function on whatever calculator you use.
For example, if you find the area between -1.96 and 1.96 in the standard normal distribution curve (mu=0, stdev=1), I'd get about .95, which means 95% of the area under the curve is between -1.96 and 1.96.
I did this on a normal distribution calculator and got .1505 (15.05%).  So the probability that 1 yellowtail selected at random is longer than 41 inches is 15.05%.
Your calculations are pretty much sound, but (41-38)/2.9 is not exactly 1.03 (it's slightly larger).  Close enough my by standards, but not totally right.
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b)  The sample mean is distributed with a mean of mu (38), and a standard deviation of $\frac{\sigma}{sqrt n}$.  N is the sample size, 32.  So do the normalcdf with mu=38, and standard deviation=0.51265241636 (0.51265241636 is 2.9/{the root of 32})
Using this on a normal distribution calculator, I got 0.  It's obviously greater than 0, but it's still really low.  I tried the Casio calculator that has super high precision, and got a whopping 2.429678289052420155163E-9 or ~.0000000002.
