Question

A professor records the time (in minutes) that it takes 16 students to complete an exam. Compute the SS, the variance, and the standard deviation assuming the 16 students constitute a population and assuming the 16 students constitute a sample. (Round your answers for variance and standard deviation to two decimal places.)
40 47 28 32
23 42 24 43
13 49 30 37
20 43 38 31
(a) the 16 students constitute a population
SS
variance
standard deviation
min

(b) the 16 students constitute a sample
SS
variance
standard deviation
min

Answer 1

Solution to 1:

Population.

$\mu = {\sum x \over N}$

$\sum x= 540$

N=16

$\mu = {540 \over 16}=33.75$

$SS = \sum {(x- \mu)}^2$

thus, SS= 1603

Variance of population

$var={SS \over N}$

$={1603/16}=100.19$ (to 2d.p)

Standard deviation

$sd={\sqrt{var}}$

$= {\sqrt{100.19}}=10.01$ (to 2 d.p)

Min

The minimum value in the population is:

13

Sample

n=16

$\bar{x} = {\sum x \over n}={540 \over 16}=33.75$

$SS={\sum(x- \bar{x})^2}$

thus, SS=1603

variance of sample

$var={SS \over n-1} = {1603 \over 15}$

=106.87 (to 2d.p)

standard deviation of sample

$sd= \sqrt{var}=\sqrt{106.87}$

=10.34 (to 2d.p)

Min

The minimum value in the sample is :

13