Question 1084646
<font color="black" face="times" size="4">Part (a) 
Correct. The fraction 1451/22 approximates to 65.9545454545454
-------------------------------------------------------
Part (b)
Notation notes:
When I say "sigma(X^2)" I mean *[Tex \Large \sum X^2] which in this case is 106639 
Similarly, "sigma(X)" means *[Tex \Large \sum X] which in this case is 1451


Using those values and n = 22, we can use the formula below to get


s = sqrt( (n*sigma(X^2)-(sigma(X))^2)/(n*(n-1)) )
s = sqrt( (22*106639-(1451)^2)/(22*(22-1)) )
s = sqrt( (22*106639-2105401)/(22*21) )
s = sqrt( (2346058-2105401)/(462) )
s = sqrt( (240657)/(462) )
s = sqrt( 520.902597402597 )
s = <font color=red>22.8232906786598</font>


which is the approximate answer to part (b)
-------------------------------------------------------
Part (c)


We're given n = 22 as the sample size.


Because n > 30 is not true and we don't know sigma (population standard deviation) we must use the T distribution.
The degrees of freedom (df) are
df = n-1
df = 22-1
df = 21


The value of xbar is approximately 65.9545454545454 as done in part (a)


Use a table <a href="http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf">such as this one</a> to locate the df = 21 row. 
I have marked this row with a red rectangle (see below)
In this df = 21 row, mark the value that is directly over top the "95%" confidence level
I marked this entire column with a blue rectangle (see below)
The red and blue rectangles intersect at the value 2.080
<img src = "https://i.imgur.com/NS3i0wy.png">
So the t critical value is t = 2.080


Earlier in part (b) we calculated the sample standard deviation to be approximately s = 22.8232906786598


We take these values (xbar = 65.9545454545454, t = 2.080, s = 22.8232906786598, n = 22) to plug into the formulas below


L = xbar - t*(s/sqrt(n))
L = 65.9545454545454 - 2.08*(22.8232906786598/sqrt(22))
L = 65.9545454545454 - 2.08*(22.8232906786598/4.69041575982343)
L = 65.9545454545454 - 2.08*(4.8659419222826)
L = 65.9545454545454 - 10.1211591983478
L = 55.8333862561976


U = xbar + t*(s/sqrt(n))
U = 65.9545454545454 + 2.08*(22.8232906786598/sqrt(22))
U = 65.9545454545454 + 2.08*(22.8232906786598/4.69041575982343)
U = 65.9545454545454 + 2.08*(4.8659419222826)
U = 65.9545454545454 + 10.1211591983478
U = 76.0757046528932


The 95% confidence interval for the mean (mu) is <font color=red>(L,U) = (55.8333862561976,76.0757046528932)</font>, which is an approximate interval.
</font>