Question

1. Women (ages 18 to 25) heights

Data Collection (Height (ft.)

1	5.0
2	5.4
3	5.4
4	5.5
5	4.11
6	5.8
7	5.1
8	5.3
9	5.4
10	5.5
11	5.7
12	5.8
13	5.10
14	5.7
15	6.3

1) Plot the first Bell Curve below:

2) What is the data value associated with a z-score of 2.1? __________________________

3) What is the data value associated with a z-score of -1.7? _________________________

4) What is the range of data values that would allow 68% of the data to fall within the mean? ______________________________________________________________________

5) What is the variance of this data set? ________________________________________

6) Are there any outliers with this data set?(Outliers refer to data points that lie beyond 3 standard deviation from the mean) _________________________________________________

Answer 1

Mean: μ = 5.41 (Using Excel function AVERAGE())

Standard deviation: σ = 0.47 (Using Excel function STDEV.P())

z-score = (x-μ)/σ

1. Using Excel function NORM.DIST(x,Mean,Std deviation, cumulative) = NORM.DIST(x,5.41,0.47,TRUE), calculate normal values

x	Normal
4.11	0.00
5	0.19
5.1	0.25
5.1	0.25
5.3	0.41
5.4	0.49
5.4	0.49
5.4	0.49
5.5	0.58
5.5	0.58
5.7	0.73
5.7	0.73
5.8	0.80
5.8	0.80
6.3	0.97

Select the above two columns, go to Insert, choose Scatter Plot with Smooth Lines.

2. z-score of 2.1 = (2.1-5.41)/0.47 = -7.04

3. z-score of -1.7 = (-1.7-5.41)/0.47 = -15.13

4. Range of data values that would allow 68% of the data to fall within the mean (Using Emperical rule): μ ± 1σ = 5.41 ± 0.47 = (4.94, 5.88)

5. Variance = σ2 = 0.47*0.47 = 0.22

6. For Outliers,

μ ± 3σ = 5.41 ± 3*0.47 = 5.41 ± 1.42 = (3.89, 6.83)

Values falling outside (3.89, 6.83) are outliers. So, there are no outliers.