The following random sample from a population whose values were normally distributed was collected.10 8 11 11The 95% confidence interval for μ is:a. 8.00 to 10.00b. 7.75 to 12.25c. 9.75 to 10.75d. 8.52 to 10.98

B

Step-by-step explanation:

First to find the sample mean

Xbar= summation x/n= 40/4= 10

The sample standard deviation is:

S= √summation(x-xbar)²/ n-1

=√6/3= 1.4142

The degrees of freedom

df= N-1= 4-1= 3

For 95% confidence level, critical value of t

t= 3.182

The 95% confidence interval is:

=Xbar plus or minus t' *8/√n

= 10 plus or minus 3.182(1.4142/√4)

= 10plus or minus 2.25

So 7.75 or 12.25

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)