Louis Math Project

GAC010 Assessment Event 2: Project-Statistics

Statistics

Student Name: Louis Chen
Teacher Name:Ian
Due Date: June 3 2015
Word Count:499

GAC010 AE2

Louis

ID

1. Create a dot plot to see the distribution of the data.
10

8

5

3

0
150

155

160

165

170

175

180

185

190

2. List data in order
157,160,162,165,166,167,167,167,168,169,170,171,172,173,174,174,176,178,180,182
3. What’s the Range of data:
Range=Highest-Lowest=182-157=25(cm)
4. Plot the data in a Histogram and describe the distribution
7

Frequency

5

4

2

0

157.5

162.5

167.5

172.5

177.5

182.5

Heights
The distribution is symmetrical. It has a normal shape. It has a slight skew to the right.

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Cumulative Frequency Percentage

5. Create a cumulative frequency curve and describe its shape.
100%

75%

50%

25%

0%

157.5

162.5

167.5

172.5

177.5

182.5

Heights
S-shaped curve. typical of a normal distribution
6. Use the ogive to estimate the median and quartiles.
1st quartile:
Q1 about 167cm
Median:
about 170 cm
3rd quartile:
Q3 About 174cm
Estimated from the graph.
7. What is the inter-quartile range?
Inter-quartile range:
IQR=Q3-Q1=174-166=7cm

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8. Box plot

160

165

170

175

180

9. Find the mean and standard deviation (sd)
!

(X − µ )2
= 6.5
!σ =
20

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10. Calculate the Pearson measure of Skewness (modal version), and the Quartile measure of skewness.
Are they good indicators of skew in the data? Give reasons for your answer.
Mode=167cm
Pearson Skewness=!

µ − Mo 169.9 − 167
=
≈ 0.446 σ 6.5

Quartile Skewness=!

(Q1 − X) + (Q3 − X) (167 − 170) + (174 − 170)
=
≈ 0.143
IQR
7

They are good indicators. Both measures indicate a positive skew, though the histogram seems to be little negatively skewed.
11. Which of the three measures of central tendency (mean, median, mode) is the best choice for this data? Give a reason for your answer.
I think mean can be the best choice. Since the data has no extreme values, nor a large skew. And mean accounts for all datas.
12. What are (mean – 1 sd) and (mean + 1 sd)? What percentage of the data is between them?
?−? = 169.9−6.5 = 163.4cm
?+? = 169.9+6.5 = 176.4cm
There are 14 data between 163.4 and 166.4
The percentage of the data: 14/20=70%
13. What are (mean – 2 sd) and (mean +2 sd)? What percentage of the data is between them?
?−2?=169.9-13= 156.9cm
?+2?=169.9+13= 182.9cm
There are 20 data between 156.9 and 182.9
The percentage of the data: 20/20=100%
14. What are (mean – 3 sd) and (mean + 3 sd)? What percentage of the data is between them?
?−3?=169.9−19.5= 150.4cm
?+3?=169.9+19.5= 189.4cm
There are 20 data
The percentage of the data: 20/20=100%

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15. Find the value in the data that is closest to z = 0.5.

X−µ σ !Z =

X close to 0.5*6.5+169.9=173.15
So X= 173
16. What percentage of the data are between z = –0.5 and 0.5?
!Z =

X−µ σ X1=Z?+? = -0.5*6.5+169.9= 166.65cm
X2=Z?+? = 0.5*6.5+169.9=173.15cm
The percentage of the data:
9/20=0.45=45%
17. The middle 50% of the data covers Q1 to Q3
So 167 ≤X≤ 174
! Z1 =

X − µ 167 − 169.9
=
≈ −0.45 σ 6.5

! Z2 =

X − µ 174 − 169.9
=
≈ 0.63 σ 6.5

Z range = 0.63-(-0.45)=1.08
18.If you chose a person from this sample randomly, what is the probability that they would be taller than
1 sd above the mean?
?=169.9
?=6.5
So, ?+?=169.9+6.5=176.4 x≥176.4 ??→ 178,180,182.
P (x ≥176.4) = 3 / 20 × 100% = 15%

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