I just need help finding out what the sample standard deviat

I just need help finding out what the sample standard deviation is. If possible please provide formula.

How much do wild mountain lions weigh? Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains gave the following weights (pounds):

Assume that the population of x values has an approximately normal distribution.

(a) Use a calculator with mean and sample standard deviation keys to find the sample mean weight x and sample standard deviation s. (Round your answers to one decimal place.)

(b) Find a 75% confidence interval for the population average weight of all adult mountain lions in the specified region. (Round your answers to one decimal place.)

Solution

a)
s.d = 31.6459
b)
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=90.66
Standard deviation( sd )=31.65
Sample Size(n)=6
Confidence Interval = [ 90.66 ± Z a/2 ( 31.65/ Sqrt ( 6) ) ]
= [ 90.66 - 1.15 * (12.92) , 90.66 + 1.15 * (12.92) ]
= [ 75.8,105.52 ]