Multiple Z-tests and independence I'm currently working through this textbook question:
"The length of three javelin throws is normally distributed, with: $$ \mu = 253, \sigma = 8.4 $$
What is the probability that the longest of his three throws is 270 ft or more?
The basic z-test is easy enough: $$ \frac{X -\mu}\sigma = \frac{270-253}{8.4} = 2.0238, P(Z>2.0238) = 0.0217$$ but I'm not sure if the bold part changes anything: I'm assuming that these throws are independent, in which case P(X>270) is the same for all throws, regardless of whether it was the longest, right? Or are additional steps required to account for the fact that this is the longest throw?
 A: Are the distributions different?

but I'm not sure if the bold part changes anything

When I get stuck figuring out whether or not two distributions are the same, I find it very helpful to simulate it and check.
Let's simulate the distribution for one javelin throw, or one draw from your normal distribution. I'm going to use Python:
from scipy.stats import norm
from numpy import sqrt
from matplotlib.pyplot import hist

# Create the normal distribution with given mean and variance
dist = norm(loc=253, scale=sqrt(8.4))

# Randomly sample 100_000 javelin throws (or draws) from the distribution
throws = dist.rvs(size=100_000)

# Make a plot
hist(throws, density=True, bins=100);


And now we want our javelin thrower to make three throws and we want to see the distribution of the longest of those three throws. To sample three throws and keep the longest one we can go max(dist.rvs(3)) and we can do this 100,000 times to create a second distribution:
longest_throw = [max(dist.rvs(3)) for _ in range(100_000)]

hist(throws, density=True, bins=100)
hist(longest_throw, density=True, bins=100, alpha=0.5)

We can show this new distribution in orange:

And indeed we find that the distribution of the longest of 3 throws is different to the distribution of one throw.
You can see that the new distribution has a higher mean. This makes sense because we are keeping the largest of three draws. We should expect at least the mean of the distribution to be higher, because we are only keeping the largest values.
Hint

Or are additional steps required to account for the fact that this is the longest throw?

You are almost there. You've got the probability that 1 throw is greater than 270 ft. How could you use that to answer this question?
Here's a hint that I find to be extremely helpful when solving probabilistic questions.
The question
What is the probability that the longest of his three throws is 270 ft or more?

is almost the same as
What is the probability that the longest of his three throws is 270 ft or less?

If $p$ is the answer to the first question, then $1 - p$ is the answer to the second question.
P.S.
In your equation, have you put in the wrong mean ($\mu$) and the wrong standard deviation ($\sigma = \sqrt{\sigma^2}$)?
