Confidence Intervals: strikeouts by Am. League leader

Over the last 15 Major League Baseball seasons, the mean # of strikeouts by the American League leader is 258.5. Assuming that the # of strikeouts by the league leader is normally distributed & the standard deviation for all seasons in all leagues is 34.9, find the 82% Confidence Interval & the 93% Confidence Interval for the mean

I get that the 82% would be 1-.82=.18 & 93% would be 1-.93=.07 and that n=15, x-bar=258.5, & st. dev=34.9. But I'm confused about how to solve. I know ^p = x/n so would that be 258.5/15? If someone could point me in the right direction, I'd appreciate it.

x = x-bar

sd = st. dev. of the mean = st. dev. / sqrt(n)

Pr(x - z(sd) < x < x + z(sd)) = 0.82

You know x.
You can calcualte sd.

Look up z in a table or have your calculator tell you what it is.

Should you be usiung a t-distribution?