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Statistics, STD2 S1 2016 HSC 29c

The ages of members of a dance class are shown in the back-to-back stem-and-leaf plot.
 

2ug-2016-hsc-q29_2
 

Pat claims that the women who attend the dance class are generally older than the men.

Is Pat correct? Justify your answer by referring to the median and skewness of the two sets of data.  (3 marks)

Show Answers Only

`text(Women:)`

`text(The median is 55 in a data)`

`text(set that is negatively skewed.)`

`text(Men:)`

`text(The median is 45 in a data)`

`text(set that is positively skewed.)`

`:.\ text(Pat is correct.)`

Show Worked Solution

`text(Women:)`

♦ Mean mark 44%.

`text(The median is 55 in a data)`

`text(set that is negatively skewed.)`

`text(Men:)`

`text(The median is 45 in a data)`

`text(set that is positively skewed.)`

`:.\ text(Pat is correct.)`

Statistics, STD2 S1 2007 HSC 24d

Barry constructed a back-to-back stem-and-leaf plot to compare the ages of his students.
 

 

  1. Write a brief statement that compares the distribution of the ages of males and females from this set of data.   (1 mark)

  2. What is the mode of this set of data?   (1 mark)

  3. Liam decided to use a grouped frequency distribution table to calculate the mean age of the students at Barry’s Ballroom Dancing Studio. 

     

    For the age group 30 - 39 years, what is the value of the product of the class centre and the frequency?   (2 marks)

  4. Liam correctly calculated the mean from the grouped frequency distribution table to be 39.5.

     

    Caitlyn correctly used the original data in the back-to-back stem-and-leaf plot and calculated the mean to be 38.2. 

     

    What is the reason for the difference in the two answers?   (1 mark)

Show Answers Only
  1. `text(More males attend than females and a higher proportion)`
    `text(of those are younger males, with the distribution being)`
    `text(positively skewed. Female attendees are generally older)`
    `text(and have a negatively skewed distribution.)`
  2. `text(Mode) = 64\ \ \ text{(4 times)}`
  3. `172.5`
  4. `text(The difference in the answers is due to the class)`
  5. `text(centres used in group frequency tables distorting)`
  6. `text(the mean value from the exact data.)`
Show Worked Solution
i. `text(More males attend than females and a higher proportion)`
  `text(of those are younger males, with the distribution being)`
  `text(positively skewed. Female attendees are generally older)`
  `text(and have a negatively skewed distribution.)`

 

ii. `text(Mode) = 64\ \ \ text{(4 times)}`

 

iii. `text(Class centre)` `= (30 + 39)/2`
    `= 34.5`
  `text(Frequency) = 5`

 
`:.\ text(Class centre) xx text(frequency)`

`= 34.5 xx 5`

`= 172.5`
 

iv. `text(The difference in the answers is due to the class)`
  `text(centres used in group frequency tables distorting)`
  `text(the mean value from the exact data.)`

Statistics, STD2 S1 2011 HSC 25d

Data was collected from 30 students on the number of text messages they had sent in the previous 24 hours. The set of data collected is displayed.
 

2UG 2011 25d

  1. What is the outlier for this set of data? (1 mark)

  2. What is the interquartile range of the data collected from the female students? (1 mark)

Show Answers Only
  1. `71`
  2. `9`
Show Worked Solution

i.   `text(Outlier is 71)`

♦♦ Mean mark 34%
COMMENT: Ensure you can quickly and accurately find quartile values using stem and leaf graphs!

ii.   `text{Lower quartile = 9   (4th female data point)}`

`text{Upper quartile = 20   (11th female data point)}`

`:.\ text{Interquartile range (female)}=20-11=9`

Statistics, STD2 S1 2010 HSC 16 MC

This back-to-back stem-and-leaf plot displays the test results for a class of 26 students.
 

2010 Q16 MC  
 

What is the median test result for the class?

  1.    `44`
  2.    `46`
  3.    `48`
  4.    `49`
Show Answers Only

`B`

Show Worked Solution
♦♦ Mean mark 35%

`text(26 results given in the data)`

  `=>text(Median is average of)\ 13^text(th)\ text(and)\ 14^text(th)`

`:.\ text(Median)` `=(45+47)/2`
  `=46`

`=>B`