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Statistics, STD1 S1 2020 HSC 24

  1. The ages in years, of ten people at the local cinema last Saturday afternoon are shown.

\(38 \ \ 25 \ \ 38 \ \ 46 \ \ 55 \ \ 68 \ \ 72 \ \ 55 \ \ 36 \ \ 38\)

  1. The mean of this dataset is 47.1 years.
  2. How many of the ten people were aged between the mean age and the median age?  (2 marks)

  3. On Wednesday, ten people all aged 70 went to this same cinema.
  4. Would the standard deviation of the age dataset from Wednesday be larger than, smaller than or equal to the standard deviation of the age dataset given in part (a)? Briefly explain your answer without performing any calculations.  (2 marks)

Show Answers Only

a.     \(1\)

b.    \(\text{Standard deviation is a measure of how much the}\)

\(\text{ages of individuals differ from the mean age of the group.}\)
 

\(\Rightarrow\ \text{Standard deviation of Wednesday’s group would be}\)

\(\text{less as the mean is 70 and everyone’s age is 70.}\)

Show Worked Solution

a.     \(\text{Reorder ages in ascending order:}\)

    \(25, 36, 38, 38, 38, 46, 55 , 55, 68, 72\)

\(\text{Median} = \dfrac{\text{5th + 6th}}{2} = \dfrac{38 + 46}{2} = 42\)

\(\therefore\ \text{People with age between 42 − 47.1 = 1}\)

♦ Mean mark (a) 39%.

 
b.    \(\text{Standard deviation is a measure of how much the}\)

\(\text{ages of individuals differ from the mean age of the group.}\)
 

\(\Rightarrow\ \text{Standard deviation of Wednesday’s group would be}\)

\(\text{less as the mean is 70 and everyone’s age is 70.}\)

♦♦♦ Mean mark (b) 20%.

Statistics, STD2 S1 2017 HSC 27a

Jamal surveyed eight households in his street. He asked them how many kilolitres (kL) of water they used in the last year. Here are the results.

`220, 105, 101, 450, 37, 338, 151, 205`

  1. Calculate the mean of this set of data.  (1 mark)

  2. What is the standard deviation of this set of data, correct to one decimal place?  (1 mark)

Show Answers Only
  1. `200.875`
  2. `127.4\ \ text{(1 d.p.)}`
Show Worked Solution
i.   `text(Mean)` `= (220 + 105 + 101 + 450 + 37 + 338 + 151 + 205) ÷ 8`
    `= 200.875`
♦ Mean mark part (ii) 47%.
IMPORTANT: The population standard deviation is required here.

 

ii.   `text(Std Dev)` `= 127.357…\ \ text{(by calc)}`
    `= 127.4\ \ text{(1 d.p.)}`

Statistics, STD2 S1 2015 HSC 27d

In a small business, the seven employees earn the following wages per week:

\(\$300, \ \$490, \ \$520, \ \$590, \ \$660, \ \$680, \ \$970\)

  1.  Is the wage of $970 an outlier for this set of data? Justify your answer with calculations.  (3 marks)

  2.  Each employee receives a $20 pay increase.

     

     What effect will this have on the standard deviation?  (1 mark)

Show Answers Only

i.    \(\text{See Worked Solutions.} \)

ii.    \(\text{The standard deviation will remain the same.}\)

Show Worked Solution

i.    \(300, 490, 520, 590, 660, 680, 970\)

\(\text{Median}\) \(= 590\)
\(Q_1\) \(= 490\)
\(Q_3\) \(= 680\)
\(IQR\) \(= 680-490 = 190\)

 

\(\text{Outlier if \$970 is greater than:} \)

\(Q_3 + 1.5 x\times IQR = 680 + 1.5 \times 190 = \$965 \) 

\(\therefore\ \text{The wage \$970 per week is an outlier.}\)

♦ Mean mark (i) 39%.


ii.    \(\text{All values increase by \$20, but so too does the mean.} \)

\(\text{Therefore the spread about the new mean will not change} \)

\(\text{and therefore the standard deviation will remain the same.} \)

Statistics, STD2 S1 2005 HSC 27d

Nine students were selected at random from a school, and their ages were recorded.

\begin{array} {|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Ages} \rule[-1ex]{0pt}{0pt} \\
\hline
\rule{0pt}{2.5ex} \ \ \ \text{12     11     16} \ \ \  \rule[-1ex]{0pt}{0pt} \\ \rule{0pt}{2.5ex} \text{14     16     15} \rule[-1ex]{0pt}{0pt} \\ \rule{0pt}{2.5ex} \text{14     15     14} \rule[-1ex]{0pt}{0pt} \\
\hline
\end{array}

  1. What is the sample standard deviation, correct to two decimal places?   (2 marks)

  2. Briefly explain what is meant by the term standard deviation.   (1 mark)

Show Answers Only
  1. `text{1.69  (to 2 d.p.)}`
  2. `text(Standard deviation is a measure of how much)`

     

    `text(members of a data group differ from the mean)`

     

    `text(value of the group)`

Show Worked Solution

i.  `text(Sample standard deviation)`

`= 1.6914…\ text{(by calculator)}`

`= 1.69\ \ \ text{(to 2 d.p.)}`

 

ii.  `text(Standard deviation is a measure of how much)`

`text(members of a data group differ from the mean)`

`text(value of the group.)`

Statistics, STD2 S5 2013 HSC 29b

Ali’s class sits two Geography tests. The results of her class on the first Geography test are shown.

`58,\ \ 74,\ \ 65,\ \ 66,\ \ 73,\ \ 71,\ \ 72,\ \ 74,\ \ 62,\ \ 70`

The mean was 68.5 for the first test. 

  1. Calculate the standard deviation for the first test. Give your answer correct to one decimal place.    (1 mark)

  2. On the second Geography test, the mean for the class was 74.4 and the standard deviation was 12.4.

     

    Ali scored 62 on the first test. Calculate the mark that she needed to obtain in the second test to ensure that her performance relative to the class was maintained.   (3 marks)

Show Answers Only
  1. `5.2\ \ \ text{(to 1 d.p.)}`
  2. `text(Ali needs to score 58.9)`
Show Worked Solution
♦ Mean mark 39%
COMMENT: Make sure you are confident with this function on your calculator!
i. `sigma` `=5.2201…`
    `=5.2`  `text{(to 1 d.p.)}`

 

ii. `z =(x-mu)/sigma`
`z text{-score (1st test)}` `= (62-68.5)/5.2`
  `=-1.25`

 
`text(2nd test has)\ z text(-score of)\-1.25 :`

♦♦ Mean mark 21%.
MARKER’S COMMENT: When “performance relative to the class is maintained”, `z text(-scores)` are the same in each test.
`-1.25` `= (x-74.4)/12.4`
`x-74.4` `=-15.5`
`x` `=58.9`

 

`:.\ text(Ali needs to score 58.9)`