Given that μ=0, and σ=1, P(z>c)=0.1093
thus
P(z<c)=1-0.1093=0.8907
hence the value of c will be:
z-score=(c-μ)/σ
thus
1.23=(c-0)/1
c=1.23
By determining the length of TV using TV^2=15^2+10^2-2(15)(10)cos80, and then determining the value of x using 15^2=TV^2+10^2-2(TV)(10)cosx.
The answer in this question is 97.
0.20 SD = 1.96 SD / sqrt(n)
n = (1.96 / .200)^2
n = 96.04
Which is rounded Up to 97
The number of observations within the data set must be greater than or equal to the quantity of 97.